2008 AIME II 详解 | LIVE by Po-Shen Loh

1.

令 $\begin{aligned} N &= 100^2 + 99^2 - 98^2 - 97^2 \\ &\quad {}+ 96^2 + \cdots + 4^2 \\ &\quad {}+ 3^2 - 2^2 - 1^2 \end{aligned}\text{，}$ 其中加号和减号每两个一组交替出现。求 $N$ 除以 $1000$ 的余数。

Let $\begin{aligned} N &= 100^2 + 99^2 - 98^2 - 97^2 \\ &\quad {}+ 96^2 + \cdots + 4^2 \\ &\quad {}+ 3^2 - 2^2 - 1^2, \end{aligned}$ where the additions and subtractions alternate in pairs. Find the remainder when $N$ is divided by $1000.$

答案

知识点：平方差配对与分组求和

难度评级：1890

小提示：

将每个带正号的平方项与比它小两个单位的带负号平方项配对，并用 $a^2 - b^2 = (a+b)(a-b)$ ，其中 $a - b = 2$

Pair each positive square with the negative square two below it, and factor $a^2 - b^2 = (a+b)(a-b)$ with $a - b = 2$

大提示：

每四个连续项分为一组；以 $(4k)^2$ 结尾的一组可化简为 $32k - 12$

Group four consecutive terms at a time; the block ending in $(4k)^2$ simplifies to $32k - 12$

解答：

每四项分成一组。对 $k = 1, 2, \ldots, 25$ ，以 $(4k)^2$ 结尾的一组为 $\begin{aligned} &(4k)^2 + (4k-1)^2 \\ &\quad {}- (4k-2)^2 - (4k-3)^2 \\ &= 2(8k - 2) + 2(8k - 4) \\ &= 32k - 12 \end{aligned}\text{，}$ 这里两次用了平方差公式 $a^2 - b^2 = (a + b)(a - b)$ ，且 $a - b = 2$ 。

对 $k = 1$ 到 $25$ ，求和， $\begin{aligned} N &= 32 \cdot \frac{25 \cdot 26}{2} - 12 \cdot 25 \\ &= 10400 - 300 = 10100 \end{aligned}\text{，}$ 所以 $N$ 除以 $1000$ 的余数是 $100$ 。

Group the terms four at a time. For $k = 1, 2, \ldots, 25,$ the block ending at $(4k)^2$ is $\begin{aligned} &(4k)^2 + (4k-1)^2 \\ &\quad {}- (4k-2)^2 - (4k-3)^2 \\ &= 2(8k - 2) + 2(8k - 4) \\ &= 32k - 12, \end{aligned}$ using the difference of squares $a^2 - b^2 = (a + b)(a - b)$ with $a - b = 2$ twice.

Summing over $k = 1$ to $25,$ $\begin{aligned} N &= 32 \cdot \frac{25 \cdot 26}{2} - 12 \cdot 25 \\ &= 10400 - 300 = 10100, \end{aligned}$ so the remainder when $N$ is divided by $1000$ is $100.$

2.

鲁道夫以恒定速度骑车，并且每骑完一英里就在里程点休息五分钟。詹妮弗也以恒定速度骑车，她的速度是鲁道夫的四分之三，但詹妮弗每骑完两英里才休息五分钟。詹妮弗和鲁道夫同时开始骑车，并且恰好同时到达 $50$ 英里标记处。问他们共用了多少分钟？

Rudolph bikes at a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the $50$ -mile mark at exactly the same time. How many minutes has it taken them?

答案

知识点：路程、速度与时间分式方程

难度评级：2020

小提示：

两人到达终点时都不再休息，所以鲁道夫休息 $49$ 次，詹妮弗休息 $24$ 次

Neither rider takes a break upon arriving, so Rudolph takes $49$ breaks and Jennifer takes $24$

大提示：

设鲁道夫的速度为每分钟 $r$ 英里，令 $\frac{50}{r} + 245$ 与 $\frac{200}{3r} + 120$ 相等

With Rudolph’s rate $r$ miles per minute, equate $\frac{50}{r} + 245$ and $\frac{200}{3r} + 120$

解答：

设鲁道夫每分钟骑 $r$ 英里。他在第 $1$ 英里到第 $49$ 英里后休息，所以总时间为 $\frac{50}{r} + 49 \cdot 5 = \frac{50}{r} + 245$ 分钟。詹妮弗每分钟骑 $\frac{3r}{4}$ 英里，并在第 $2, 4, \ldots, 48$ 英里后休息，所以她的总时间为 $\frac{50}{\frac{3r}{4}} + 24 \cdot 5 = \frac{200}{3r} + 120$ 分钟。

令两人的时间相等，得到 $\frac{50}{r} + 245 = \frac{200}{3r} + 120\text{，}$ 所以 $\begin{aligned} 125 &= \frac{200 - 150}{3r} \\ &= \frac{50}{3r} \end{aligned}\text{，}$ 因此 $r = \frac{2}{15}$ 。共同用时为 $\frac{50}{\frac{2}{15}} + 245 = 375 + 245 = 620$ 分钟。

Let Rudolph bike at $r$ miles per minute. He rests after each of miles $1$ through $49,$ so his total time is $\frac{50}{r} + 49 \cdot 5 = \frac{50}{r} + 245$ minutes. Jennifer bikes at $\frac{3r}{4}$ miles per minute and rests after each of miles $2, 4, \ldots, 48,$ so her total time is $\frac{50}{\frac{3r}{4}} + 24 \cdot 5 = \frac{200}{3r} + 120$ minutes.

Setting the times equal gives $\frac{50}{r} + 245 = \frac{200}{3r} + 120,$ so $\begin{aligned} 125 &= \frac{200 - 150}{3r} \\ &= \frac{50}{3r}, \end{aligned}$ and $r = \frac{2}{15}.$ The common time is $\frac{50}{\frac{2}{15}} + 245 = 375 + 245 = 620$ minutes.

3.

一块长方体形状的奶酪尺寸为 $10$ 厘米、 $13$ 厘米、 $14$ 厘米。从这块奶酪上切下十片。每片厚 $1$ 厘米，且平行于奶酪的某一个面切下。各片不一定彼此平行。切下十片后，剩余奶酪块的体积最大可能是多少立方厘米？

A block of cheese in the shape of a rectangular solid measures $10$ cm by $13$ cm by $14$ cm. Ten slices are cut from the cheese. Each slice has a width of $1$ cm and is cut parallel to one face of the cheese. The individual slices are not necessarily parallel to each other. What is the maximum possible volume in cubic cm of the remaining block of cheese after ten slices have been cut off?

答案

知识点：算术-几何平均不等式体积最优化

难度评级：1970

小提示：

每片都会使三个维度中的一个减少 $1$ 厘米，所以剩余长方体的三边长之和为 $10 + 13 + 14 - 10 = 27$

Each slice shortens one of the three dimensions by $1$ cm, so the remaining block has dimensions summing to $10 + 13 + 14 - 10 = 27$

大提示：

正数之和固定时，它们的乘积在这些数全相等时最大

A product of positive numbers with a fixed sum is largest when the numbers are all equal

解答：

每片厚 $1$ 厘米且平行于某个面，所以每次切割后剩余奶酪仍是一个长方体，只是某个维度减少 $1$ 。若十片分别使三个维度减少 $p$ 、 $q$ 、 $r$ ，其中 $p + q + r = 10$ ，则剩余长方体尺寸为 $(10 - p) \times (13 - q) \times (14 - r)$ ，这些维度之和为 $37 - 10 = 27$ 。

由算术平均数与几何平均数不等式，和为 $27$ 的正数乘积在三者都等于 $9$ 时最大；这可以通过切下 $1$ 片，使 $10$ 厘米这一维缩短，切下 $4$ 片，使 $13$ 厘米这一维缩短，再切下 $5$ 片，使 $14$ 厘米这一维缩短来实现。最大体积为 $9^3 = 729$ 立方厘米。

Every slice is $1$ cm wide and parallel to a face, so after each cut the remaining cheese is still a rectangular block, with one dimension shortened by $1.$ If the ten slices shorten the three dimensions by $p,$ $q,$ and $r$ with $p + q + r = 10,$ the remaining block measures $(10 - p) \times (13 - q) \times (14 - r),$ and these dimensions sum to $37 - 10 = 27.$

By the AM-GM inequality, a product of positive numbers with fixed sum $27$ is greatest when all three are equal to $9,$ which is achieved by taking $1$ slice from the $10$ cm dimension, $4$ from the $13$ cm dimension, and $5$ from the $14$ cm dimension. The maximum volume is $9^3 = 729$ cubic cm.

4.

存在唯一的一组 $r$ 个非负整数 $n_1 \gt n_2 \gt \cdots \gt n_r$ ，以及唯一确定的 $r$ 个整数 $a_k$ $(1 \le k \le r)$ ，其中每个 $a_k$ 都等于 $1$ 或 $-1$ ，使得 $\begin{aligned} &a_1 3^{n_1} + a_2 3^{n_2} + \cdots + a_r 3^{n_r} \\ &= 2008 \end{aligned}\text{。}$ 求 $n_1 + n_2 + \cdots + n_r$ 。

There exist $r$ unique nonnegative integers $n_1 \gt n_2 \gt \cdots \gt n_r$ and $r$ unique integers $a_k$ $(1 \le k \le r)$ with each $a_k$ either $1$ or $-1$ such that $\begin{aligned} &a_1 3^{n_1} + a_2 3^{n_2} + \cdots + a_r 3^{n_r} \\ &= 2008. \end{aligned}$ Find $n_1 + n_2 + \cdots + n_r.$

答案

知识点：进制指数

难度评级：2350

小提示：

先把 $2008$ 写成 $3$ 进制

Start by writing $2008$ in base $3$

大提示：

把每个数字 $2$ 用公式 $2 \cdot 3^k = 3^{k+1} - 3^k$ 替换，使每个系数都变成 $1$ 或 $-1$

Replace each digit $2$ using $2 \cdot 3^k = 3^{k+1} - 3^k$ so that every coefficient becomes $1$ or $-1$

解答：

在 $3$ 进制下， $2008 = 2202101_3$ ，也就是说 $\begin{aligned} 2008 &= 2 \cdot 3^6 + 2 \cdot 3^5 + 2 \cdot 3^3 \\ &\quad {}+ 3^2 + 3^0 \end{aligned}\text{。}$ 为了把数字 $2$ 转成系数 $\pm 1$ ，使用 $2 \cdot 3^k = 3^{k+1} - 3^k$ 。两个相邻的数字 $2$ 会整齐抵消： $\begin{aligned} 2 \cdot 3^6 + 2 \cdot 3^5 &= (3^7 - 3^6) \\ &\quad {}+ (3^6 - 3^5) \\ &= 3^7 - 3^5 \end{aligned}\text{，}$ 并且 $2 \cdot 3^3 = 3^4 - 3^3$ 。

因此 $\begin{aligned} 2008 &= 3^7 - 3^5 + 3^4 - 3^3 \\ &\quad {}+ 3^2 + 3^0 \end{aligned}\text{，}$ 它有互不相同的指数和系数 $\pm 1$ ，符合要求。指数之和为 $7 + 5 + 4 + 3 + 2 + 0 = 21$ 。

In base $3,$ $2008 = 2202101_3,$ that is, $\begin{aligned} 2008 &= 2 \cdot 3^6 + 2 \cdot 3^5 + 2 \cdot 3^3 \\ &\quad {}+ 3^2 + 3^0. \end{aligned}$ To convert the digits $2$ into coefficients $\pm 1,$ use $2 \cdot 3^k = 3^{k+1} - 3^k.$ The two adjacent digits $2$ collapse neatly: $\begin{aligned} 2 \cdot 3^6 + 2 \cdot 3^5 &= (3^7 - 3^6) \\ &\quad {}+ (3^6 - 3^5) \\ &= 3^7 - 3^5, \end{aligned}$ and $2 \cdot 3^3 = 3^4 - 3^3.$

Therefore $\begin{aligned} 2008 &= 3^7 - 3^5 + 3^4 - 3^3 \\ &\quad {}+ 3^2 + 3^0, \end{aligned}$ which has distinct exponents and coefficients $\pm 1,$ as required. The sum of the exponents is $7 + 5 + 4 + 3 + 2 + 0 = 21.$

5.

在梯形 $ABCD$ 中， $\overline{BC} \parallel \overline{AD}$ ，且 $BC = 1000$ ， $AD = 2008$ 。已知 $\angle A = 37^\circ$ ， $\angle D = 53^\circ$ ，并且 $M$ 和 $N$ 分别是 $\overline{BC}$ 和 $\overline{AD}$ 的中点。求 $MN$ 的长度。

In trapezoid $ABCD$ with $\overline{BC} \parallel \overline{AD},$ let $BC = 1000$ and $AD = 2008.$ Let $\angle A = 37^\circ,$ $\angle D = 53^\circ,$ and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD},$ respectively. Find the length $MN.$

答案

知识点：梯形直角三角形中线（几何）位似

难度评级：2480

小提示：

延长腰 $AB$ 和 $DC$ 交于 $E$ ；因为 $37^\circ + 53^\circ = 90^\circ$ ，点 $E$ 处的角是直角

Extend the legs $AB$ and $DC$ to meet at $E;$ since $37^\circ + 53^\circ = 90^\circ,$ the angle at $E$ is right

大提示：

直角三角形斜边上的中线等于斜边的一半，并且 $E$ 、 $M$ 、 $N$ 共线

The median to the hypotenuse of a right triangle is half the hypotenuse, and $E,$ $M,$ $N$ are collinear

解答：

延长 $\overline{AB}$ 和 $\overline{DC}$ ，使它们交于点 $E$ 。由于 $\angle A + \angle D = 37^\circ + 53^\circ = 90^\circ$ ，三角形 $EAD$ 在 $E$ 处为直角。又因为 $\overline{BC} \parallel \overline{AD}$ ，三角形 $EBC$ 是三角形 $EAD$ 在以 $E$ 为中心的位似变换下的像，所以中点 $M$ （位于 $\overline{BC}$ 上）对应于中点 $N$ （位于 $\overline{AD}$ 上）；特别地， $E$ 、 $M$ 、 $N$ 共线。

直角三角形斜边上的中线等于斜边的一半，所以 $EN = \frac{2008}{2} = 1004$ ，且 $EM = \frac{1000}{2} = 500$ 。因此 $\begin{aligned} MN &= EN - EM \\ &= 1004 - 500 = 504 \end{aligned}\text{。}$

Extend legs $\overline{AB}$ and $\overline{DC}$ until they meet at a point $E.$ Since $\angle A + \angle D = 37^\circ + 53^\circ = 90^\circ,$ triangle $EAD$ has a right angle at $E.$ Because $\overline{BC} \parallel \overline{AD},$ triangle $EBC$ is the image of triangle $EAD$ under a homothety centered at $E,$ so the midpoint $M$ of $\overline{BC}$ maps to the midpoint $N$ of $\overline{AD};$ in particular $E,$ $M,$ and $N$ are collinear.

The median to the hypotenuse of a right triangle is half the hypotenuse, so $EN = \frac{2008}{2} = 1004$ and $EM = \frac{1000}{2} = 500.$ Therefore $\begin{aligned} MN &= EN - EM \\ &= 1004 - 500 = 504. \end{aligned}$

6.

数列 $\{a_n\}$ 定义为 $\begin{aligned} a_0 &= 1, \\ a_1 &= 1, \\ a_n &= a_{n-1} + \frac{a_{n-1}^2}{a_{n-2}} \quad (n \ge 2) \end{aligned}\text{。}$

数列 $\{b_n\}$ 定义为 $\begin{aligned} b_0 &= 1, \\ b_1 &= 3, \\ b_n &= b_{n-1} + \frac{b_{n-1}^2}{b_{n-2}} \quad (n \ge 2) \end{aligned}\text{。}$

求 $\frac{b_{32}}{a_{32}}$ 。

The sequence $\{a_n\}$ is defined by $\begin{aligned} a_0 &= 1, \\ a_1 &= 1, \\ a_n &= a_{n-1} + \frac{a_{n-1}^2}{a_{n-2}} \quad (n \ge 2). \end{aligned}$

The sequence $\{b_n\}$ is defined by $\begin{aligned} b_0 &= 1, \\ b_1 &= 3, \\ b_n &= b_{n-1} + \frac{b_{n-1}^2}{b_{n-2}} \quad (n \ge 2). \end{aligned}$

Find $\frac{b_{32}}{a_{32}}.$

答案

知识点：递推阶乘裂项相消

难度评级：2460

小提示：

将递推式除以 $a_{n-1}$ 可看出比值 $\frac{a_n}{a_{n-1}}$ 每一步恰好增加 $1$

Divide the recurrence by $a_{n-1}$ to see that the ratio $\frac{a_n}{a_{n-1}}$ increases by exactly $1$ at each step

大提示：

这些比值给出 $a_n = n!$ 和 $b_n = \frac{(n+2)!}{2}$ ；再求商

The ratios give $a_n = n!$ and $b_n = \frac{(n+2)!}{2};$ form the quotient

解答：

将递推式除以 $a_{n-1}$ ，得到 $\frac{a_n}{a_{n-1}} = 1 + \frac{a_{n-1}}{a_{n-2}}\text{，}$ 所以相邻项比值每一步恰好增加 $1$ 。对 $\{a_n\}$ ，第一个比值为 $\frac{a_1}{a_0} = 1$ ，所以 $\frac{a_n}{a_{n-1}} = n$ ，从而 $a_n = n!$ 。同样的计算适用于 $\{b_n\}$ ，它的第一个比值为 $\frac{b_1}{b_0} = 3$ ，所以 $\frac{b_n}{b_{n-1}} = n + 2$ ，且 $b_n = \frac{(n+2)!}{2}$ 。

因此 $\frac{b_{32}}{a_{32}} = \frac{\frac{34!}{2}}{32!} = \frac{34 \cdot 33}{2} = 561\text{。}$

Dividing the recurrence by $a_{n-1}$ gives $\frac{a_n}{a_{n-1}} = 1 + \frac{a_{n-1}}{a_{n-2}},$ so the consecutive-term ratio increases by exactly $1$ each step. For $\{a_n\}$ the first ratio is $\frac{a_1}{a_0} = 1,$ so $\frac{a_n}{a_{n-1}} = n$ and $a_n = n!.$ The same computation applies to $\{b_n\},$ whose first ratio is $\frac{b_1}{b_0} = 3,$ so $\frac{b_n}{b_{n-1}} = n + 2$ and $b_n = \frac{(n+2)!}{2}.$

Therefore $\frac{b_{32}}{a_{32}} = \frac{\frac{34!}{2}}{32!} = \frac{34 \cdot 33}{2} = 561.$

7.

令 $r$ 、 $s$ 、 $t$ 为方程 $8x^3 + 1001x + 2008 = 0$ 的三个根。求 $(r + s)^3 + (s + t)^3 + (t + r)^3$ 。

Let $r,$ $s,$ and $t$ be the three roots of the equation $8x^3 + 1001x + 2008 = 0.$ Find $(r + s)^3 + (s + t)^3 + (t + r)^3.$

答案

知识点：韦达定理多项式立方和与立方差

难度评级：2410

小提示：

没有 $x^2$ 项，所以 $r + s + t = 0$ ，且 $r + s$ 、 $s + t$ 、 $t + r$ 分别是某个根的相反数

There is no $x^2$ term, so $r + s + t = 0$ and each of $r + s,$ $s + t,$ $t + r$ is the negative of a root

大提示：

当 $x + y + z = 0$ 时，可用恒等式 $x^3 + y^3 + z^3 = 3xyz$ ；韦达定理由常数项给出根的乘积

When $x + y + z = 0,$ the identity $x^3 + y^3 + z^3 = 3xyz$ applies; Vieta gives the product of the roots from the constant term

解答：

该三次方程没有 $x^2$ 项，所以由韦达定理 $r + s + t = 0$ 。于是 $r + s = -t$ ， $s + t = -r$ ，且 $t + r = -s$ ，所求和为 $\begin{aligned} &(-t)^3 + (-r)^3 \\ &\quad {}+ (-s)^3 \\ &= -(r^3 + s^3 + t^3) \end{aligned}\text{。}$

当 $r + s + t = 0$ 时，恒等式 $r^3 + s^3 + t^3 - 3rst$ $= (r + s + t)$ $(r^2 + s^2 + t^2 - rs - st - tr)$ 给出 $r^3 + s^3 + t^3 = 3rst$ 。由韦达定理， $rst = -\frac{2008}{8} = -251$ ，所以 $r^3 + s^3 + t^3 = -753$ ，答案为 $-(-753) = 753$ 。

The cubic has no $x^2$ term, so $r + s + t = 0$ by Vieta’s formulas. Hence $r + s = -t,$ $s + t = -r,$ and $t + r = -s,$ and the desired sum is $\begin{aligned} &(-t)^3 + (-r)^3 \\ &\quad {}+ (-s)^3 \\ &= -(r^3 + s^3 + t^3). \end{aligned}$

Whenever $r + s + t = 0,$ the identity $r^3 + s^3 + t^3 - 3rst$ $= (r + s + t)$ $(r^2 + s^2 + t^2 - rs - st - tr)$ gives $r^3 + s^3 + t^3 = 3rst.$ By Vieta’s formulas, $rst = -\frac{2008}{8} = -251,$ so $r^3 + s^3 + t^3 = -753,$ and the answer is $-(-753) = 753.$

8.

令 $a = \frac{\pi}{2008}$ 。求最小的正整数 $n$ ，使得 $\begin{aligned} &2[\cos(a)\sin(a) + \cos(4a)\sin(2a) \\ &\quad {}+ \cos(9a)\sin(3a) \\ &\quad {}+ \cdots + \cos(n^2 a)\sin(na)] \end{aligned}$ 是整数。

Let $a = \frac{\pi}{2008}.$ Find the smallest positive integer $n$ such that $\begin{aligned} &2[\cos(a)\sin(a) + \cos(4a)\sin(2a) \\ &\quad {}+ \cos(9a)\sin(3a) \\ &\quad {}+ \cdots + \cos(n^2 a)\sin(na)] \end{aligned}$ is an integer.

答案

知识点：三角恒等式裂项相消整除性

难度评级：2740

小提示：

使用积化和差公式，将每一项化为两个正弦项之差

Apply the product-to-sum identity to turn each term into a difference of two sine terms

大提示：

该和逐项抵消后化为 $\sin(n(n+1)a)$ ，它为整数当且仅当 $n(n+1)$ 是 $1004$ 的倍数

The sum telescopes to $\sin(n(n+1)a),$ which is an integer exactly when $n(n+1)$ is a multiple of $1004$

解答：

由积化和差公式， $\begin{aligned} &2\cos(k^2 a)\sin(ka) \\ &= \sin(k^2 a + ka) \\ &\quad {}- \sin(k^2 a - ka) \\ &= \sin(k(k+1)a) \\ &\quad {}- \sin((k-1)k a) \end{aligned}\text{。}$ 对 $k = 1$ 到 $n$ 求和后，各项裂项相消，只剩 $\sin(n(n+1)a) = \sin\frac{n(n+1)\pi}{2008}\text{。}$

正弦值为整数只可能是 $-1$ ， $0$ ，或 $1$ ，也就是它的角必须是 $\frac{\pi}{2}$ 的倍数。因此需要 $\frac{n(n+1)}{2008}$ 是 $\frac{1}{2}$ 的倍数，即 $n(n+1)$ 是 $1004$ 的倍数，其中 $1004 = 4 \cdot 251$ ，且 $251$ 是质数。

由于 $n$ 与 $n + 1$ 互质， $251$ 必须整除其中一个，所以 $n \ge 250$ 。当 $n = 250$ 时，乘积 $250 \cdot 251$ 不能被 $4$ 整除。当 $n = 251$ 时，乘积 $251 \cdot 252$ 能被 $4 \cdot 251 = 1004$ 整除。最小这样的 $n$ 是 $251$ 。

By the product-to-sum identity, $\begin{aligned} &2\cos(k^2 a)\sin(ka) \\ &= \sin(k^2 a + ka) \\ &\quad {}- \sin(k^2 a - ka) \\ &= \sin(k(k+1)a) \\ &\quad {}- \sin((k-1)k a). \end{aligned}$ Summing over $k = 1$ to $n,$ the terms telescope, leaving $\sin(n(n+1)a) = \sin\frac{n(n+1)\pi}{2008}.$

A sine is an integer only when it is $-1,$ $0,$ or $1,$ that is, when its argument is a multiple of $\frac{\pi}{2}.$ So we need $\frac{n(n+1)}{2008}$ to be a multiple of $\frac{1}{2},$ i.e. $n(n+1)$ is divisible by $1004,$ where $1004 = 4 \cdot 251$ and $251$ is prime.

Since $n$ and $n + 1$ are coprime, $251$ must divide one of them, so $n \ge 250.$ For $n = 250$ the product $250 \cdot 251$ is not divisible by $4.$ For $n = 251$ the product $251 \cdot 252$ is divisible by $4 \cdot 251 = 1004.$ The smallest such $n$ is $251.$

9.

一个粒子位于坐标平面上的 $(5, 0)$ 。定义粒子的一次移动为：先绕原点逆时针旋转 $\frac{\pi}{4}$ 弧度，再平移 $10$ 个单位，方向为 $x$ 轴正方向。已知粒子移动 $150$ 次后的位置为 $(p, q)$ ，求小于或等于 $|p| + |q|$ 的最大整数。

A particle is located on the coordinate plane at $(5, 0).$ Define a move for the particle as a counterclockwise rotation of $\frac{\pi}{4}$ radians about the origin followed by a translation of $10$ units in the positive $x$ -direction. Given that the particle’s position after $150$ moves is $(p, q),$ find the greatest integer less than or equal to $|p| + |q|.$

答案

知识点：复数单位根等比数列

难度评级：2840

小提示：

在复平面中，一次移动把 $z$ 变为 $\omega z + 10$ ，其中 $\omega = e^{\frac{i\pi}{4}}$

In the complex plane a move sends $z$ to $\omega z + 10,$ where $\omega = e^{\frac{i\pi}{4}}$

大提示：

迭代后，将所得等比和按每 $8$ 项一组，再利用 $\omega^8 = 1$ 化简

After iterating, group the resulting geometric sum into blocks of $8$ using $\omega^8 = 1$

解答：

将平面看作复平面，一次移动把 $z$ 变为 $\omega z + 10$ ，其中 $\omega = e^{\frac{i\pi}{4}}$ 。从 $z_0 = 5$ 开始迭代， $\begin{aligned} z_{150} &= 5\omega^{150} \\ &\quad {}+ 10 \scriptsize\left(\omega^{149} + \omega^{148} + \cdots + \omega + 1\right) \end{aligned}\text{。}$

因为 $\omega^8 = 1$ 且 $150 = 8 \cdot 18 + 6$ ，所以 $\omega^{150} = \omega^6 = -i$ 。在等比和中，每 $8$ 个连续幂之和为 $0$ ，因此这 $150$ 项化为 $\begin{aligned} &1 + \omega + \cdots + \omega^5 \\ &= -\omega^6 - \omega^7 \\ &= i - \frac{\sqrt{2}}{2} \\ &\quad {}+ \frac{\sqrt{2}}{2}i \end{aligned}\text{。}$ 因此 $\begin{aligned} z_{150} &= -5i \\ &\quad {}+ 10 \scriptsize\left(-\frac{\sqrt{2}}{2} + \left(1 + \frac{\sqrt{2}}{2}\right)i\right) \\ &= -5\sqrt{2} + \left(5 + 5\sqrt{2}\right)i \end{aligned}\text{。}$

所以 $\begin{aligned} |p| + |q| &= 5\sqrt{2} + 5 + 5\sqrt{2} \\ &= 5 + 10\sqrt{2} \\ &\approx 19.14 \end{aligned}\text{，}$ 小于或等于它的最大整数是 $19$ 。

Identify the plane with the complex plane, so a move sends $z$ to $\omega z + 10$ with $\omega = e^{\frac{i\pi}{4}}.$ Starting from $z_0 = 5$ and iterating, $\begin{aligned} z_{150} &= 5\omega^{150} \\ &\quad {}+ 10 \scriptsize\left(\omega^{149} + \omega^{148} + \cdots + \omega + 1\right). \end{aligned}$

Since $\omega^8 = 1$ and $150 = 8 \cdot 18 + 6,$ we get $\omega^{150} = \omega^6 = -i.$ In the geometric sum, every block of $8$ consecutive powers adds to $0,$ so the $150$ terms reduce to $\begin{aligned} &1 + \omega + \cdots + \omega^5 \\ &= -\omega^6 - \omega^7 \\ &= i - \frac{\sqrt{2}}{2} \\ &\quad {}+ \frac{\sqrt{2}}{2}i. \end{aligned}$ Therefore $\begin{aligned} z_{150} &= -5i \\ &\quad {}+ 10 \scriptsize\left(-\frac{\sqrt{2}}{2} + \left(1 + \frac{\sqrt{2}}{2}\right)i\right) \\ &= -5\sqrt{2} + \left(5 + 5\sqrt{2}\right)i. \end{aligned}$

Thus $\begin{aligned} |p| + |q| &= 5\sqrt{2} + 5 + 5\sqrt{2} \\ &= 5 + 10\sqrt{2} \\ &\approx 19.14, \end{aligned}$ and the greatest integer less than or equal to this is $19.$

10.

下图显示一个 $4 \times 4$ 的矩形点阵，每个点与其最近邻点相距 $1$ 个单位。

定义一条增长路径为点阵中一列互不相同的点，并且序列中相邻两点之间的距离严格递增。令 $m$ 为增长路径可能包含的最大点数，令 $r$ 为恰好包含 $m$ 个点的增长路径条数。求 $mr$ 。

The diagram below shows a $4 \times 4$ rectangular array of points, each of which is $1$ unit away from its nearest neighbors.

Define a growing path to be a sequence of distinct points of the array with the property that the distance between consecutive points of the sequence is strictly increasing. Let $m$ be the maximum possible number of points in a growing path, and let $r$ be the number of growing paths consisting of exactly $m$ points. Find $mr.$

答案

知识点：格点距离公式极端原理分类讨论

难度评级：3060

小提示：

两个点阵点之间的距离平方为 $a^2 + b^2$ ，其中 $0 \le a, b \le 3$ 且不同时为零；这只有 $9$ 个不同取值

The squared distance between two array points is $a^2 + b^2$ with $0 \le a, b \le 3$ not both zero, which takes only $9$ distinct values

大提示：

从最长的一步开始反向构造最长路径：最后两个点必须是相对的角点，此后几乎每个更早的点都被迫确定

Build maximal paths backwards from the longest step: the last two points must be opposite corners, and nearly every earlier point is then forced

解答：

点阵中两点之间的距离平方为 $a^2 + b^2$ ，其中 $a$ 和 $b$ 是坐标差，均属于 $\{0, 1, 2, 3\}$ 且不同时为零。可能的值为 $1, 2, 4, 5, 8, 9, 10, 13, 18$ ，只有 $9$ 个值，所以增长路径最多有 $10$ 个点；若有 $10$ 个点，则必须按递增顺序使用全部九种距离。把这些点标为 $P_1, \ldots, P_{10}$ ，使 $P_1 P_2 = 1$ 且 $P_9 P_{10} = \sqrt{18}$ 。

$\sqrt{18}$ 只能由相对的角点实现，所以有 $4$ 个有序选择 $(P_{10}, P_9)$ 。接下来， $P_8 P_9 = \sqrt{13}$ 留下 $2$ 种 $P_8$ 的选择，即 $P_{10}$ 的两个相邻点，它们关于主对角线对称。从这里开始，距离 $\sqrt{10}, 3, \sqrt{8}, \sqrt{5}, 2, \sqrt{2}$ 会唯一决定 $P_7, P_6, \ldots, P_2$ （对 $P_7$ ，另一个角点选择不可行，因为下一步需要的 $P_6$ 会与 $P_9$ 或 $P_{10}$ 重合）。最后， $P_1$ 必须与 $P_2$ 相距 $1$ ，而后者的邻点中有 $3$ 个尚未使用。下面显示其中一条路径。

因此 $m = 10$ ， $r = 4 \cdot 2 \cdot 3 = 24$ ，所以 $mr = 240$ 。

The squared distance between two points of the array is $a^2 + b^2,$ where $a$ and $b$ are the coordinate differences, each in $\{0, 1, 2, 3\}$ and not both zero. The possible values are $1, 2, 4, 5, 8, 9, 10, 13, 18$ — only $9$ values — so a growing path has at most $10$ points, and a path with $10$ points must use all nine distances in increasing order. Label its points $P_1, \ldots, P_{10}$ so that $P_1 P_2 = 1$ and $P_9 P_{10} = \sqrt{18}.$

Since $\sqrt{18}$ is realized only by opposite corners, there are $4$ ordered choices of $(P_{10}, P_9).$ Next, $P_8 P_9 = \sqrt{13}$ leaves $2$ choices for $P_8,$ the two neighbors of $P_{10},$ symmetric across the main diagonal. From there the distances $\sqrt{10}, 3, \sqrt{8}, \sqrt{5}, 2, \sqrt{2}$ force $P_7, P_6, \ldots, P_2$ uniquely (for $P_7$ the alternative corner choice fails because the point needed next for $P_6$ would coincide with $P_9$ or $P_{10}$ ). Finally $P_1$ must be at distance $1$ from $P_2,$ and $3$ of its neighbors are unused. One of the resulting paths is shown below.

Hence $m = 10$ and $r = 4 \cdot 2 \cdot 3 = 24,$ so $mr = 240.$

11.

在三角形 $ABC$ 中， $AB = AC = 100$ ，且 $BC = 56$ 。圆 $P$ 的半径为 $16$ ，并与 $\overline{AC}$ 和 $\overline{BC}$ 相切。圆 $Q$ 与圆 $P$ 外切，并与 $\overline{AB}$ 和 $\overline{BC}$ 相切。圆 $Q$ 没有任何点在 $\triangle ABC$ 外。圆 $Q$ 的半径可表示为 $m - n\sqrt{k}$ ，其中 $m$ 、 $n$ 、 $k$ 是正整数，且 $k$ 是若干不同质数的乘积。求 $m + nk$ 。

In triangle $ABC,$ $AB = AC = 100,$ and $BC = 56.$ Circle $P$ has radius $16$ and is tangent to $\overline{AC}$ and $\overline{BC}.$ Circle $Q$ is externally tangent to circle $P$ and is tangent to $\overline{AB}$ and $\overline{BC}.$ No point of circle $Q$ lies outside of $\triangle ABC.$ The radius of circle $Q$ can be expressed in the form $m - n\sqrt{k},$ where $m,$ $n,$ and $k$ are positive integers and $k$ is the product of distinct primes. Find $m + nk.$

答案

知识点：相切圆坐标几何角平分线三角恒等式

难度评级：2990

小提示：

取 $B = (0,0)$ ， $C = (56,0)$ ， $A = (28,96)$ ；两个圆心都在角平分线上，并且 $\tan\frac{B}{2} = \tan\frac{C}{2} = \frac{3}{4}$

Place $B = (0,0),$ $C = (56,0),$ $A = (28,96);$ both centers lie on angle bisectors, and $\tan\frac{B}{2} = \tan\frac{C}{2} = \frac{3}{4}$

大提示：

两个圆心为 $\left(56 - \frac{64}{3},\, 16\right)$ 和 $\left(\frac{4q}{3},\, q\right)$ ；令它们之间的距离等于 $q + 16$

The centers are $\left(56 - \frac{64}{3},\, 16\right)$ and $\left(\frac{4q}{3},\, q\right);$ set the distance between them equal to $q + 16$

解答：

取 $B = (0, 0)$ ， $C = (56, 0)$ ；从 $A$ 向底边作的高为 $\sqrt{100^2 - 28^2} = 96$ ，所以 $A = (28, 96)$ 。于是 $\sin B = \frac{24}{25}$ ， $\cos B = \frac{7}{25}$ ，且 $\begin{aligned} \tan\frac{B}{2} &= \frac{\sin B}{1 + \cos B} \\ &= \frac{3}{4} = \tan\frac{C}{2} \end{aligned}\text{。}$ 一个半径为 $r$ ，且与 $\overline{BC}$ 和一条斜边相切的圆，其圆心位于从相应底角顶点出发的角平分线上，高度为 $r$ ，与该顶点的水平距离为 $\frac{r}{\tan(\frac{C}{2})} = \frac{4r}{3}$ 。因此 $P = \left(56 - \frac{64}{3},\, 16\right)$ ，且 $Q = \left(\frac{4q}{3},\, q\right)$ ，其中 $q$ 是圆 $Q$ 的半径。

外切意味着 $PQ = q + 16$ ： $\begin{aligned} &\left(\frac{104 - 4q}{3}\right)^2 + (16 - q)^2 \\ &= (16 + q)^2 \end{aligned}\text{。}$ 由于 $(16 + q)^2 - (16 - q)^2 = 64q$ ，这变为 $(104 - 4q)^2 = 576q$ ，即 $(26 - q)^2 = 36q$ ，化简得 $q^2 - 88q + 676 = 0$ ，所以 $q = 44 \pm 6\sqrt{35}$ 。

根 $44 + 6\sqrt{35} \approx 79.5$ 会使圆 $Q$ 伸出三角形外，所以 $q = 44 - 6\sqrt{35}$ 。因此 $m = 44$ ， $n = 6$ ， $k = 35 = 5 \cdot 7$ ，得到 $m + nk = 44 + 210 = 254$ 。

Place $B = (0, 0)$ and $C = (56, 0);$ the altitude from $A$ has length $\sqrt{100^2 - 28^2} = 96,$ so $A = (28, 96).$ Then $\sin B = \frac{24}{25},$ $\cos B = \frac{7}{25},$ and $\begin{aligned} \tan\frac{B}{2} &= \frac{\sin B}{1 + \cos B} \\ &= \frac{3}{4} = \tan\frac{C}{2}. \end{aligned}$ A circle of radius $r$ tangent to $\overline{BC}$ and to a slanted side has its center on the bisector from that base vertex, at height $r$ and horizontal distance $\frac{r}{\tan(\frac{C}{2})} = \frac{4r}{3}$ from the vertex. Thus $P = \left(56 - \frac{64}{3},\, 16\right)$ and $Q = \left(\frac{4q}{3},\, q\right),$ where $q$ is the radius of circle $Q.$

External tangency means $PQ = q + 16:$ $\begin{aligned} &\left(\frac{104 - 4q}{3}\right)^2 + (16 - q)^2 \\ &= (16 + q)^2. \end{aligned}$ Since $(16 + q)^2 - (16 - q)^2 = 64q,$ this becomes $(104 - 4q)^2 = 576q,$ i.e. $(26 - q)^2 = 36q,$ which simplifies to $q^2 - 88q + 676 = 0,$ so $q = 44 \pm 6\sqrt{35}.$

The root $44 + 6\sqrt{35} \approx 79.5$ would make circle $Q$ extend outside the triangle, so $q = 44 - 6\sqrt{35}.$ Here $m = 44,$ $n = 6,$ and $k = 35 = 5 \cdot 7,$ giving $m + nk = 44 + 210 = 254.$

12.

有两根可区分的旗杆，还有 $19$ 面旗，其中 $10$ 面是完全相同的蓝旗， $9$ 面是完全相同的绿旗。令 $N$ 为使用所有旗帜的可区分排列数，要求每根旗杆上至少有一面旗，并且任一旗杆上没有两面绿旗相邻。求 $N$ 除以 $1000$ 的余数。

There are two distinguishable flagpoles, and there are $19$ flags, of which $10$ are identical blue flags, and $9$ are identical green flags. Let $N$ be the number of distinguishable arrangements using all of the flags in which each flagpole has at least one flag and no two green flags on either pole are adjacent. Find the remainder when $N$ is divided by $1000.$

答案

知识点：有限制的排列组合补集计数范德蒙德卷积

难度评级：3060

小提示：

在一根有 $b$ 面蓝旗和 $g$ 面绿旗的旗杆上，绿旗必须占据蓝旗形成的 $b + 1$ 个空隙中的不同空隙：有 $\binom{b+1}{g}$ 种

On a pole with $b$ blue and $g$ green flags, the greens must occupy distinct gaps among the $b + 1$ gaps determined by the blues: $\binom{b+1}{g}$ ways

大提示：

用范德蒙德恒等式对 $\binom{b+1}{g}\binom{11-b}{9-g}$ 关于 $g$ 求和，再减去某根旗杆为空的排列

Sum $\binom{b+1}{g}\binom{11-b}{9-g}$ over $g$ using Vandermonde’s identity, then subtract the arrangements that leave a pole empty

解答：

假设第一根旗杆得到 $b$ 面蓝旗和 $g$ 面绿旗，第二根得到剩余的 $10 - b$ 面蓝旗和 $9 - g$ 面绿旗。在一根有 $b$ 面蓝旗的旗杆上，绿旗必须占据蓝旗周围 $b + 1$ 个空隙中的不同空隙，有 $\binom{b+1}{g}$ 种。暂时不考虑每根旗杆非空的要求，总数为 $\begin{aligned} &\sum_{b=0}^{10} \sum_{g=0}^{9} \binom{b+1}{g} \\ &\quad {}\cdot \binom{11-b}{9-g} \\ &= \sum_{b=0}^{10} \binom{12}{9} \\ &= 11 \cdot 220 = 2420 \end{aligned}\text{，}$ 其中内层和由范德蒙德恒等式化简，因为 $(b + 1) + (11 - b) = 12$ 。

使一根旗杆为空的排列，是把全部 $19$ 面旗放在另一根旗杆上；每选择一根空旗杆，都有 $\binom{11}{9} = 55$ 种。因此 $N = 2420 - 2 \cdot 55 = 2310\text{，}$ 所以 $N$ 除以 $1000$ 的余数为 $310$ 。

Suppose the first pole gets $b$ blue and $g$ green flags, the second the remaining $10 - b$ blue and $9 - g$ green. On a pole with $b$ blue flags, the green flags must occupy distinct gaps among the $b + 1$ gaps around the blues, in $\binom{b+1}{g}$ ways. Temporarily ignoring the requirement that each pole be nonempty, the total is $\begin{aligned} &\sum_{b=0}^{10} \sum_{g=0}^{9} \binom{b+1}{g} \\ &\quad {}\cdot \binom{11-b}{9-g} \\ &= \sum_{b=0}^{10} \binom{12}{9} \\ &= 11 \cdot 220 = 2420, \end{aligned}$ where the inner sum collapses by Vandermonde’s identity, since $(b + 1) + (11 - b) = 12.$

The arrangements that leave a pole empty put all $19$ flags on one pole, in $\binom{11}{9} = 55$ ways for each choice of pole. Hence $N = 2420 - 2 \cdot 55 = 2310,$ and the remainder when $N$ is divided by $1000$ is $310.$

13.

复平面中，一个以原点为中心的正六边形，其每对相对边之间相距一单位。其中一对边平行于虚轴。令 $R$ 为六边形外部区域，并令 $S = \left\{\tfrac{1}{z} \mid z \in R\right\}$ 。则 $S$ 的面积形如 $a\pi + \sqrt{b}$ ，其中 $a$ 和 $b$ 是正整数。求 $a + b$ 。

A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \left\{\tfrac{1}{z} \mid z \in R\right\}.$ Then the area of $S$ has the form $a\pi + \sqrt{b},$ where $a$ and $b$ are positive integers. Find $a + b.$

答案

知识点：复数变换扇形对称性

难度评级：3370

小提示：

写 $w = \frac{1}{z}$ ，半平面 $\mathrm{Re}\,z \gt \frac{1}{2}$ 恰好映射到以 $1$ 为圆心、半径为 $1$ 的圆盘

Writing $w = \frac{1}{z},$ the half-plane $\mathrm{Re}\,z \gt \frac{1}{2}$ maps exactly onto the disk of radius $1$ centered at $1$

大提示：

$S$ 是以六次单位根为圆心的六个单位圆盘的并；利用对称性，在一个 $60^\circ$ 扇形内求面积，即一个 $120^\circ$ 扇形加两个三角形

$S$ is the union of six unit disks centered at the sixth roots of unity; by symmetry find the area inside one $60^\circ$ wedge as a $120^\circ$ sector plus two triangles

解答：

六边形的边到原点的距离为 $\frac{1}{2}$ ，其中一条边在直线 $\mathrm{Re}\,z = \frac{1}{2}$ 上，所以 $R$ 是把半平面 $\mathrm{Re}\,z \gt \frac{1}{2}$ 旋转 $60^\circ$ 的倍数得到的六个半平面的并。若 $w = u + vi = \frac{1}{z}$ ，则 $\mathrm{Re}\,z = \mathrm{Re}\,\frac{1}{w} = \frac{u}{u^2 + v^2} \gt \frac{1}{2}$ 等价于 $u^2 + v^2 \lt 2u$ ，即 $(u - 1)^2 + v^2 \lt 1$ 。因此每个半平面映射成一个开单位圆盘， $S$ 是以六次单位根为圆心的六个单位圆盘的并。

把平面分成六个 $60^\circ$ 扇形，分界射线的方向角为 $30^\circ + 60^\circ k$ 。由对称性，在每个扇形内， $S$ 与圆心位于该扇形中的那个圆盘重合。方向角为 $\pm 30^\circ$ 的射线与圆 $|w - 1| = 1$ 交于 $\left(\frac{3}{2}, \pm\frac{\sqrt{3}}{2}\right)$ ，所以 $S$ 在该扇形内的部分包括两个三角形。每个三角形的顶点为 $0$ 、圆心 $1$ 和其中一个交点，是两腰长为 $1$ 、顶角为 $120^\circ$ 的等腰三角形，面积为 $\frac{\sqrt{3}}{4}$ ；此外还有这两个交点之间的 $120^\circ$ 圆扇形，面积为 $\frac{\pi}{3}$ 。

因此每个六十度扇形贡献 $\frac{\pi}{3} + \frac{\sqrt{3}}{2}$ ，总面积为 $\begin{aligned} 6\left(\frac{\pi}{3} + \frac{\sqrt{3}}{2}\right) &= 2\pi + 3\sqrt{3} \\ &= 2\pi + \sqrt{27} \end{aligned}\text{。}$ 因而 $a = 2$ ， $b = 27$ ，所以 $a + b = 29$ 。

The hexagon’s sides lie at distance $\frac{1}{2}$ from the origin, with one side on the line $\mathrm{Re}\,z = \frac{1}{2},$ so $R$ is the union of the six half-planes obtained by rotating $\mathrm{Re}\,z \gt \frac{1}{2}$ by multiples of $60^\circ.$ If $w = u + vi = \frac{1}{z},$ then $\mathrm{Re}\,z = \mathrm{Re}\,\frac{1}{w} = \frac{u}{u^2 + v^2} \gt \frac{1}{2}$ is equivalent to $u^2 + v^2 \lt 2u,$ i.e. $(u - 1)^2 + v^2 \lt 1.$ So each half-plane maps onto an open unit disk, and $S$ is the union of six unit disks centered at the sixth roots of unity.

Cut the plane into six $60^\circ$ wedges by the rays at angles $30^\circ + 60^\circ k;$ by symmetry, within each wedge $S$ coincides with the disk whose center lies in that wedge. The rays at $\pm 30^\circ$ meet the circle $|w - 1| = 1$ at $\left(\frac{3}{2}, \pm\frac{\sqrt{3}}{2}\right),$ so the piece of $S$ in that wedge consists of two triangles with vertices at $0,$ the center $1,$ and one of these points — each isosceles with two sides $1$ and apex angle $120^\circ,$ area $\frac{\sqrt{3}}{4}$ — together with the $120^\circ$ sector of the disk between them, area $\frac{\pi}{3}.$

Each wedge therefore contributes $\frac{\pi}{3} + \frac{\sqrt{3}}{2},$ and the total area is $\begin{aligned} 6\left(\frac{\pi}{3} + \frac{\sqrt{3}}{2}\right) &= 2\pi + 3\sqrt{3} \\ &= 2\pi + \sqrt{27}. \end{aligned}$ Thus $a = 2,$ $b = 27,$ and $a + b = 29.$

14.

设 $a$ 和 $b$ 为正实数，且 $a \ge b$ 。令 $\rho$ 为 $\frac{a}{b}$ 的最大可能值，使得方程组 $\begin{aligned} a^2 + y^2 &= b^2 + x^2 \\ &= (a - x)^2 + (b - y)^2 \end{aligned}$ 存在解 $(x, y)$ ，且满足 $0 \le x \lt a$ 、 $0 \le y \lt b$ 。则 $\rho^2$ 可表示为分数 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

Let $a$ and $b$ be positive real numbers with $a \ge b.$ Let $\rho$ be the maximum possible value of $\frac{a}{b}$ for which the system of equations $\begin{aligned} a^2 + y^2 &= b^2 + x^2 \\ &= (a - x)^2 + (b - y)^2 \end{aligned}$ has a solution $(x, y)$ satisfying $0 \le x \lt a$ and $0 \le y \lt b.$ Then $\rho^2$ can be expressed as a fraction $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

知识点：等边三角形距离公式三角学最优化

难度评级：3270

小提示：

相等的量是距离平方：取 $D = (0,b)$ 、 $E = (x,0)$ 、 $F = (a,b-y)$ ，它们位于一个 $a \times b$ 矩形中，则三角形 $DEF$ 是等边三角形

The equal quantities are squared distances: with $D = (0,b),$ $E = (x,0),$ $F = (a,b-y)$ in an $a \times b$ rectangle, triangle $DEF$ is equilateral

大提示：

令 $\theta = \angle ADE$ ，等边条件迫使 $\frac{a}{b} = \frac{\cos(30^\circ - \theta)} {\cos\theta}$ ，它随 $\theta$ 增大，而 $y \ge 0$ 将 $\theta$ 限制到至多 $30^\circ$

With $\theta = \angle ADE,$ equal sides force $\frac{a}{b} = \frac{\cos(30^\circ - \theta)} {\cos\theta},$ which increases in $\theta,$ while $y \ge 0$ caps $\theta$ at $30^\circ$

解答：

画一个顶点为 $A = (0, 0)$ 、 $B = (a, 0)$ 、 $C = (a, b)$ 、 $D = (0, b)$ 的矩形，并令 $E = (x, 0)$ 在 $\overline{AB}$ 上， $F = (a, b - y)$ 在 $\overline{BC}$ 上。那么 $DE^2 = b^2 + x^2$ ， $DF^2 = a^2 + y^2$ ，且 $EF^2 = (a - x)^2 + (b - y)^2$ ，所以方程组正是说三角形 $DEF$ 是等边三角形，而约束条件保证 $E$ 与 $F$ 在这两条边上。

令 $\theta = \angle ADE$ ，则 $x = b\tan\theta$ ，且 $DE = \frac{b}{\cos\theta}$ 。因为 $\angle EDF = 60^\circ$ ，而 $D$ 处的矩形角是 $90^\circ$ ，所以 $\angle CDF = 30^\circ - \theta$ ，从而 $y = a\tan(30^\circ - \theta)$ ，且 $DF = \frac{a}{\cos(30^\circ - \theta)}$ 。令 $DE = DF$ ，得到 $\begin{aligned} \frac{a}{b} &= \frac{\cos(30^\circ - \theta)}{\cos\theta} \\ &= \cos 30^\circ + \sin 30^\circ \tan\theta \end{aligned}\text{，}$ 它随 $\theta$ 增大而增大。条件 $x \ge 0$ 给出 $\theta \ge 0$ ，而 $y \ge 0$ 迫使 $\theta \le 30^\circ$ 。

因此最大值在 $\theta = 30^\circ$ 时取得，此时 $\frac{a}{b} = \frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}} = \frac{2}{\sqrt{3}}$ ，并可由 $y = 0$ 和 $x = \frac{b}{\sqrt{3}} \lt a$ 实现。于是 $\rho^2 = \frac{4}{3}$ ，且 $m + n = 4 + 3 = 7$ 。

Draw the rectangle with vertices $A = (0, 0),$ $B = (a, 0),$ $C = (a, b),$ $D = (0, b),$ and let $E = (x, 0)$ on $\overline{AB}$ and $F = (a, b - y)$ on $\overline{BC}.$ Then $DE^2 = b^2 + x^2,$ $DF^2 = a^2 + y^2,$ and $EF^2 = (a - x)^2 + (b - y)^2,$ so the system says exactly that triangle $DEF$ is equilateral, with the constraints keeping $E$ and $F$ on those two sides.

Let $\theta = \angle ADE,$ so $x = b\tan\theta$ and $DE = \frac{b}{\cos\theta}.$ Since $\angle EDF = 60^\circ$ and the corner angle at $D$ is $90^\circ,$ we get $\angle CDF = 30^\circ - \theta,$ so $y = a\tan(30^\circ - \theta)$ and $DF = \frac{a}{\cos(30^\circ - \theta)}.$ Setting $DE = DF$ gives $\begin{aligned} \frac{a}{b} &= \frac{\cos(30^\circ - \theta)}{\cos\theta} \\ &= \cos 30^\circ + \sin 30^\circ \tan\theta, \end{aligned}$ which is increasing in $\theta.$ The requirement $x \ge 0$ gives $\theta \ge 0,$ while $y \ge 0$ forces $\theta \le 30^\circ.$

The maximum is therefore at $\theta = 30^\circ,$ where $\frac{a}{b} = \frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}} = \frac{2}{\sqrt{3}},$ attained with $y = 0$ and $x = \frac{b}{\sqrt{3}} \lt a.$ Hence $\rho^2 = \frac{4}{3},$ and $m + n = 4 + 3 = 7.$

15.

求满足以下条件的最大整数 $n$ ： $\mathrm{(i)}$ $n^2$ 可表示为两个连续立方数之差； $\mathrm{(ii)}$ $2n + 79$ 是一个完全平方数。

Find the largest integer $n$ satisfying the following conditions: $\mathrm{(i)}$ $n^2$ can be expressed as the difference of two consecutive cubes; $\mathrm{(ii)}$ $2n + 79$ is a perfect square.

答案

知识点：丢番图方程完全平方数平方差模运算

难度评级：3160

小提示：

条件 $\mathrm{(i)}$ 表示 $n^2 = 3m^2 + 3m + 1$ ，可整理为 $(2n - 1)(2n + 1) = 3(2m + 1)^2$

Condition $\mathrm{(i)}$ says $n^2 = 3m^2 + 3m + 1,$ which rearranges to $(2n - 1)(2n + 1) = 3(2m + 1)^2$

大提示：

两个因子互质，且模 $3$ 可知 $2n - 1$ 必须是完全平方数；然后 $2n + 79 = d^2$ 给出一个等于 $80$ 的平方差

The two factors are coprime, and working modulo $3$ shows $2n - 1$ must be a perfect square; then $2n + 79 = d^2$ yields a difference of squares equal to $80$

解答：

条件 $\mathrm{(i)}$ 表示 $n^2 = (m + 1)^3 - m^3$ $= 3m^2 + 3m + 1$ 对某个整数 $m$ 成立。两边乘以 $4$ 并整理，得 $4n^2 - 1 = 12m^2 + 12m + 3$ ，即 $(2n - 1)(2n + 1) = 3(2m + 1)^2\text{。}$ 左边两个因子是相邻奇数，因此互质，所以一个是完全平方数，另一个是 $3$ 倍的完全平方数。若 $2n - 1 = 3k^2$ ，则 $2n + 1 \equiv 2 \pmod 3$ 会是一个完全平方数，这是不可能的。因此 $2n - 1 = k^2$ ，其中 $k$ 为奇数。

写 $k = 2a + 1$ ，得到 $n = 2a^2 + 2a + 1$ 。条件 $\mathrm{(ii)}$ 表示 $2n + 79 = 4a^2 + 4a + 81 = d^2$ ，所以 $\begin{aligned} &(d - 2a - 1) \\ &\quad {}\cdot (d + 2a + 1) \\ &= d^2 - (2a + 1)^2 \\ &= 80 \end{aligned}\text{。}$ 两个因子奇偶性相同，所以都为偶数：因子对 $(2, 40)$ 、 $(4, 20)$ 、 $(8, 10)$ 分别给出 $2a + 1 = 19$ 、 $8$ 、 $1$ ，其中奇数值给出 $a = 9$ （于是 $n = 181$ ）和 $a = 0$ （于是 $n = 1$ ）。

对 $n = 181$ ：确有 $181^2 = 32761 = 105^3 - 104^3$ （此时 $2n + 1 = 363 = 3 \cdot 11^2$ ，符合要求），且 $2n + 79 = 441 = 21^2$ 。所以满足条件的最大 $n$ 是 $181$ 。

Condition $\mathrm{(i)}$ says $n^2 = (m + 1)^3 - m^3$ $= 3m^2 + 3m + 1$ for some integer $m.$ Multiplying by $4$ and rearranging, $4n^2 - 1 = 12m^2 + 12m + 3,$ i.e. $(2n - 1)(2n + 1) = 3(2m + 1)^2.$ The factors on the left are consecutive odd numbers, hence coprime, so one of them is a perfect square and the other is $3$ times a square. If $2n - 1 = 3k^2,$ then $2n + 1 \equiv 2 \pmod 3$ would be a perfect square, which is impossible. Hence $2n - 1 = k^2$ with $k$ odd.

Writing $k = 2a + 1$ gives $n = 2a^2 + 2a + 1.$ Condition $\mathrm{(ii)}$ says $2n + 79 = 4a^2 + 4a + 81 = d^2,$ so $\begin{aligned} &(d - 2a - 1) \\ &\quad {}\cdot (d + 2a + 1) \\ &= d^2 - (2a + 1)^2 \\ &= 80. \end{aligned}$ The two factors have the same parity, so both are even: the pairs $(2, 40),$ $(4, 20),$ $(8, 10)$ give $2a + 1 = 19,$ $8,$ $1,$ of which the odd values yield $a = 9$ (so $n = 181$ ) and $a = 0$ (so $n = 1$ ).

For $n = 181:$ indeed $181^2 = 32761 = 105^3 - 104^3$ (here $2n + 1 = 363 = 3 \cdot 11^2,$ as required), and $2n + 79 = 441 = 21^2.$ So the largest such $n$ is $181.$