2004 AIME II 真题与在线限时测验

计时

3:00:00

1.

圆的一条弦垂直于一条半径，且交点是这条半径的中点。该弦把圆分成两个区域，较大区域面积与较小区域面积的比可表示为 $\frac{a\pi + b\sqrt{c}}{d\pi - e\sqrt{f}}$ ，其中 $a$ ， $b$ ， $c$ ， $d$ ， $e$ ， $f$ 都是正整数， $a$ 与 $e$ 互质，且 $c$ 和 $f$ 都不被任何质数的平方整除。求 $a \cdot b \cdot c \cdot d \cdot e \cdot f$ 除以 $1000$ 的余数。

A chord of a circle is perpendicular to a radius at the midpoint of the radius. The ratio of the area of the larger of the two regions into which the chord divides the circle to the smaller can be expressed in the form $\frac{a\pi + b\sqrt{c}}{d\pi - e\sqrt{f}},$ where $a,$ $b,$ $c,$ $d,$ $e,$ and $f$ are positive integers, $a$ and $e$ are relatively prime, and neither $c$ nor $f$ is divisible by the square of any prime. Find the remainder when the product $a \cdot b \cdot c \cdot d \cdot e \cdot f$ is divided by $1000.$

答案

答案：592

知识点：扇形弦圆面积

难度评级：2050

小提示：

令半径为 $2$ 。弦到圆心的距离为 $1$ ，所以它所对的圆心角为 $120^\circ$ 。

Take the radius to be $2;$ the chord then lies at distance $1$ from the center, so it subtends a $120^\circ$ central angle

大提示：

较小区域是圆盘的三分之一减去面积为 $\sqrt{3}$ 的三角形；较大区域是圆盘的三分之二再加上这个三角形。

The smaller region is a third of the disk minus a triangle of area $\sqrt{3};$ the larger is two-thirds of the disk plus that triangle

解答：

按比例缩放，使半径为 $2$ 。弦到圆心的距离为 $1$ ，所以连到弦两端的半径各与被平分的半径成 $60^\circ$ 角，两条端点半径形成的圆心角为 $120^\circ$ 。它们截出的等腰三角形面积为 $\frac{1}{2} \cdot 2 \cdot 2 \sin 120^\circ = \sqrt{3}$ ，整个圆盘面积为 $4\pi$ 。

较小区域是 $120^\circ$ 扇形减去该三角形，面积为 $\frac{4\pi}{3} - \sqrt{3}$ ；较大区域是剩余部分，面积为 $\frac{8\pi}{3} + \sqrt{3}$ 。比值为 $\frac{\frac{8\pi}{3} + \sqrt{3}}{\frac{4\pi}{3} - \sqrt{3}} = \frac{8\pi + 3\sqrt{3}}{4\pi - 3\sqrt{3}}\text{，}$ 因而 $(a, b, c, d, e, f) = (8, 3, 3, 4, 3, 3)$ 。

乘积为 $8 \cdot 3 \cdot 3 \cdot 4 \cdot 3 \cdot 3 = 2592$ ，除以 $1000$ 的余数为 $592$ 。

Scale so the radius is $2.$ The chord lies at distance $1$ from the center, so each radius to an endpoint of the chord makes a $60^\circ$ angle with the bisected radius, and the two endpoint radii form a central angle of $120^\circ.$ The isosceles triangle they cut off has area $\frac{1}{2} \cdot 2 \cdot 2 \sin 120^\circ = \sqrt{3},$ and the whole disk has area $4\pi.$

The smaller region is the $120^\circ$ sector minus the triangle, $\frac{4\pi}{3} - \sqrt{3},$ and the larger region is the rest, $\frac{8\pi}{3} + \sqrt{3}.$ The ratio is $\frac{\frac{8\pi}{3} + \sqrt{3}}{\frac{4\pi}{3} - \sqrt{3}} = \frac{8\pi + 3\sqrt{3}}{4\pi - 3\sqrt{3}},$ which has the required form with $(a, b, c, d, e, f) = (8, 3, 3, 4, 3, 3).$

The product is $8 \cdot 3 \cdot 3 \cdot 4 \cdot 3 \cdot 3 = 2592,$ whose remainder upon division by $1000$ is $592.$

2.

一个罐子里有 $10$ 颗红色糖果和 $10$ 颗蓝色糖果。Terry 随机取出两颗糖果，然后 Mary 从剩下的糖果中随机取出两颗。已知他们取到的颜色组合（不考虑顺序）相同的概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

A jar has $10$ red candies and $10$ blue candies. Terry picks two candies at random, then Mary picks two of the remaining candies at random. Given that the probability that they get the same color combination, irrespective of order, is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

答案

答案：441

知识点：无放回抽样组合分类讨论

难度评级：2180

小提示：

他们的组合相同只可能是两人都取到两红、两人都取到两蓝，或两人都各取一红一蓝。

They match only if both draw two reds, both draw two blues, or both draw one candy of each color

大提示：

对两人都取两红的情形，概率为 $\frac{\binom{10}{2}}{\binom{20}{2}} \cdot \frac{\binom{8}{2}}{\binom{18}{2}}$ 。

Mary draws from what Terry left: for two reds each, the probability is $\frac{\binom{10}{2}}{\binom{20}{2}} \cdot \frac{\binom{8}{2}}{\binom{18}{2}}$

解答：

颜色组合相同恰好发生在两人都取两红、两人都取两蓝，或两人都取一红一蓝。Terry 取两红的概率为 $\frac{\binom{10}{2}}{\binom{20}{2}} = \frac{45}{190} = \frac{9}{38}$ 。此后剩下 $8$ 颗红色糖果和 $10$ 颗蓝色糖果，所以 Mary 取两红的概率为 $\frac{\binom{8}{2}}{\binom{18}{2}} = \frac{28}{153}$ 。该情形的概率为 $\frac{9}{38} \cdot \frac{28}{153} = \frac{14}{323}$ ；由对称性，两人都取两蓝的情形也有同样的概率 $\frac{14}{323}$ 。

对各取一红一蓝的情形，Terry 取到这种组合的概率为 $\frac{10 \cdot 10}{\binom{20}{2}} = \frac{10}{19}$ ，此后两种颜色各剩 $9$ 颗，Mary 取到这种组合的概率为 $\frac{9 \cdot 9}{\binom{18}{2}} = \frac{9}{17}$ ，所以该情形概率为 $\frac{10}{19} \cdot \frac{9}{17} = \frac{90}{323}$ 。

总概率为 $\frac{14 + 14 + 90}{323} = \frac{118}{323}$ 。因为 $118 = 2 \cdot 59$ ， $323 = 17 \cdot 19$ ，该分数已最简，所以 $m + n = 118 + 323 = 441$ 。

The combinations match exactly when both draw two reds, both draw two blues, or both draw one candy of each color. The probability that Terry draws two reds is $\frac{\binom{10}{2}}{\binom{20}{2}} = \frac{45}{190} = \frac{9}{38},$ after which $8$ reds and $10$ blues remain, so Mary draws two reds with probability $\frac{\binom{8}{2}}{\binom{18}{2}} = \frac{28}{153}.$ That case has probability $\frac{9}{38} \cdot \frac{28}{153} = \frac{14}{323},$ and by symmetry two blues each is also $\frac{14}{323}.$

For mixed draws, Terry succeeds with probability $\frac{10 \cdot 10}{\binom{20}{2}} = \frac{10}{19},$ leaving $9$ of each color, and Mary with probability $\frac{9 \cdot 9}{\binom{18}{2}} = \frac{9}{17},$ for $\frac{10}{19} \cdot \frac{9}{17} = \frac{90}{323}.$

The total is $\frac{14 + 14 + 90}{323} = \frac{118}{323}.$ Since $118 = 2 \cdot 59$ and $323 = 17 \cdot 19,$ the fraction is in lowest terms, and $m + n = 118 + 323 = 441.$

3.

一个实心长方体由 $N$ 个全等的、边长为 $1$ 厘米的小立方体面贴面粘成。当从能看见它三个面的方向观察时，恰有 $231$ 个边长为 $1$ 厘米的小立方体看不见。求 $N$ 的最小可能值。

A solid rectangular block is formed by gluing together $N$ congruent $1$ -cm cubes face to face. When the block is viewed so that three of its faces are visible, exactly $231$ of the $1$ -cm cubes cannot be seen. Find the smallest possible value of $N.$

答案

答案：384

知识点：长方体质因数分解最优化

难度评级：2110

小提示：

看不见的立方体正好是不接触这三个可见面的那些，它们形成一个 $(p-1) \times (q-1) \times (r-1)$ 的长方体。

The unseen cubes are exactly those behind the three visible faces: they form a $(p-1) \times (q-1) \times (r-1)$ block

大提示：

将 $231 = 3 \cdot 7 \cdot 11$ 分解为三个正整数的乘积，逐一比较对应长方体的体积。

Factor $231 = 3 \cdot 7 \cdot 11$ as a product of three positive integers in every possible way and compare the resulting volumes

解答：

设长方体尺寸为 $p \times q \times r$ 。一个小立方体看不见，当且仅当它不接触三个可见面中的任何一个，所以看不见的小立方体形成 $(p-1) \times (q-1) \times (r-1)$ 的块。因此 $(p-1)(q-1)(r-1) = 231$ $= 3 \cdot 7 \cdot 11$ 。

将 $231$ 写成三个正整数乘积的方式为 $3 \cdot 7 \cdot 11$ 、 $1 \cdot 3 \cdot 77$ 、 $1 \cdot 7 \cdot 33$ 、 $1 \cdot 11 \cdot 21$ 、 $1 \cdot 1 \cdot 231$ 。对应长方体为 $4 \times 8 \times 12$ 、 $2 \times 4 \times 78$ 、 $2 \times 8 \times 34$ 、 $2 \times 12 \times 22$ 、 $2 \times 2 \times 232$ ，体积分别为 $384$ ， $624$ ， $544$ ， $528$ ， $928$ 。

最小值为 $N = 384$ 。

Let the block measure $p \times q \times r.$ A cube is hidden exactly when it touches none of the three visible faces, so the hidden cubes form a $(p-1) \times (q-1) \times (r-1)$ block, giving $(p-1)(q-1)(r-1) = 231$ $= 3 \cdot 7 \cdot 11.$

The ways to write $231$ as a product of three positive integers are $3 \cdot 7 \cdot 11,$ $1 \cdot 3 \cdot 77,$ $1 \cdot 7 \cdot 33,$ $1 \cdot 11 \cdot 21,$ and $1 \cdot 1 \cdot 231,$ giving blocks $4 \times 8 \times 12,$ $2 \times 4 \times 78,$ $2 \times 8 \times 34,$ $2 \times 12 \times 22,$ and $2 \times 2 \times 232,$ with volumes $384,$ $624,$ $544,$ $528,$ and $928.$

The smallest is $N = 384.$

4.

小于 $10{,}000$ 的正整数中，有多少个至多含有两种不同的数字？

How many positive integers less than $10{,}000$ have at most two different digits?

答案

答案：927

知识点：数字分类讨论乘法原理

难度评级：2300

小提示：

一位数和两位数都符合条件；对 $3$ 位数和 $4$ 位数，可按两种数字分别占据哪些位置来计数。

Every number with one or two digits qualifies; count $3$ - and $4$ -digit numbers by which positions hold which of two digit values

大提示：

一个 $4$ 位数若首位数字为 $a$ ，另一种数字为 $b \ne a$ ，则后三位中出现第二种数字的非空位置模式有 $2^3 - 1 = 7$ 种；每种模式有 $9 \cdot 9$ 种数字取法，再加上 $9$ 个各位相同的数。

A $4$ -digit number with leading digit $a$ and second value $b \ne a$ has $2^3 - 1 = 7$ patterns, each in $9 \cdot 9$ ways; add the $9$ numbers whose digits are all equal

解答：

共有 $99$ 个小于 $100$ 的正整数，它们都符合条件。符合条件的 $3$ 位数要么是各位相同的数（共 $9$ 个），要么使用首位数字 $a \ge 1$ 和另一个数字 $b \ne a$ ，且第二种数字出现在后两位的某个非空位置集合中。这样的模式有 $2^2 - 1 = 3$ 种，每种有 $9 \cdot 9$ 种取值（先有 $9$ 种 $a$ 的选择，再有 $9$ 种 $b$ 的选择），所以三位数共有 $9 + 3 \cdot 81 = 252$ 个。

同理，符合条件的 $4$ 位数要么是各位相同的数（共 $9$ 个），要么让 $b \ne a$ 出现在后三位的某个非空位置集合中；这样的模式有 $2^3 - 1 = 7$ 种，每种有 $9 \cdot 9$ 种数字取法，所以四位数共有 $9 + 7 \cdot 81 = 576$ 个。

总数为 $99 + 252 + 576 = 927$ 。

All $99$ positive integers below $100$ qualify. A qualifying $3$ -digit number is either a number whose digits are all equal ( $9$ of them) or uses a leading digit $a \ge 1$ together with a second value $b \ne a$ in some of the last two positions: $2^2 - 1 = 3$ patterns, each realized in $9 \cdot 9$ ways ( $9$ choices for $a,$ then $9$ for $b$ ), for $9 + 3 \cdot 81 = 252$ numbers.

Similarly a qualifying $4$ -digit number has all its digits equal ( $9$ possibilities) or has $b \ne a$ appearing in a nonempty subset of the last three positions: $2^3 - 1 = 7$ patterns, each in $9 \cdot 9$ ways, for $9 + 7 \cdot 81 = 576$ numbers.

The total is $99 + 252 + 576 = 927.$

5.

为完成一项大工程，雇用了 $1000$ 名工人，人数刚好足以按期完成。所有工人在完成前四分之一工程时都留在工地，所以前四分之一按时完成。之后解雇了 $100$ 名工人，于是第二个四分之一延误完成。又解雇了 $100$ 名工人，于是第三个四分之一完成时进度进一步落后。已知所有工人的工作效率相同，在完成四分之三工程后，为使整个工程按期或提前完成，至少还必须在剩下的 $800$ 名工人之外再雇用多少名工人？

In order to complete a large job, $1000$ workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then $100$ workers were laid off, so the second quarter of the work was completed behind schedule. Then an additional $100$ workers were laid off, so the third quarter of the work was completed still further behind schedule. Given that all workers work at the same rate, what is the minimum number of additional workers, beyond the $800$ workers still on the job at the end of the third quarter, that must be hired after three-quarters of the work has been completed so that the entire project can be completed on schedule or before?

答案

答案：766

知识点：速率不等式

难度评级：2390

小提示：

选取时间单位，使整个工程计划用 $4$ 个单位完成；第一个四分之一工程正好用 $1$ 个单位。

Choose the time unit so the whole job is scheduled for $4$ units; the first quarter then takes exactly $1$ unit

大提示：

前三个四分之一共用时 $1 + \frac{10}{9} + \frac{5}{4}$ ，剩下最后四分之一只有 $\frac{23}{36}$ 个时间单位。

The first three quarters take $1 + \frac{10}{9} + \frac{5}{4}$ units, leaving $\frac{23}{36}$ of a unit for the final quarter

解答：

取时间单位，使 $1000$ 名工人完成四分之一工程需要 $1$ 个单位；总工期为 $4$ 个单位。 $900$ 名工人完成第二个四分之一需要 $\frac{10}{9}$ 个单位， $800$ 名工人完成第三个四分之一需要 $\frac{10}{8} = \frac{5}{4}$ 个单位。已用时间为 $1 + \frac{10}{9} + \frac{5}{4} = \frac{121}{36}\text{，}$ 剩余时间为 $4 - \frac{121}{36} = \frac{23}{36}$ 个单位。

最后四分之一需要 $1000$ 个工人时间单位的劳动量，所以总工人数 $w$ 必须满足 $w \cdot \frac{23}{36} \ge 1000$ ，即 $w \ge \frac{36000}{23} \approx 1565.2$ ，因此至少需要 $1566$ 名工人。

工地上还剩 $800$ 名工人，所以至少还要雇用 $1566 - 800 = 766$ 名工人。

Measure time so that $1000$ workers complete a quarter of the job in $1$ unit; the schedule allows $4$ units in all. With $900$ workers the second quarter takes $\frac{10}{9}$ units, and with $800$ workers the third quarter takes $\frac{10}{8} = \frac{5}{4}$ units. The time used so far is $1 + \frac{10}{9} + \frac{5}{4} = \frac{121}{36},$ leaving $4 - \frac{121}{36} = \frac{23}{36}$ of a unit for the last quarter.

The last quarter requires $1000$ worker-units of labor, so the workforce $w$ must satisfy $w \cdot \frac{23}{36} \ge 1000,$ that is $w \ge \frac{36000}{23} \approx 1565.2,$ so at least $1566$ workers are needed.

Since $800$ remain on the job, at least $1566 - 800 = 766$ additional workers must be hired.

6.

三只聪明的猴子分一堆香蕉。第一只猴子从堆中拿走一些香蕉，自己留下其中的四分之三，并把剩下的平均分给另外两只。第二只猴子从堆中拿走一些香蕉，自己留下其中的四分之一，并把剩下的平均分给另外两只。第三只猴子拿走堆中剩下的香蕉，自己留下其中的十二分之一，并把剩下的平均分给另外两只。已知每次分香蕉时每只猴子都得到整数个香蕉，且最后第一、第二、第三只猴子得到的香蕉数之比为 $3 : 2 : 1$ 。香蕉总数的最小可能值是多少？

Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio $3 : 2 : 1,$ what is the least possible total for the number of bananas?

答案

答案：408

知识点：方程组比与比例丢番图方程

难度评级：2400

小提示：

设三只猴子分别拿走 $8x$ 、 $8y$ 、 $24z$ 个香蕉，这样每次留下和分给别人都自动是整数。

Let the monkeys take $8x,$ $8y,$ and $24z$ bananas, so every keep-and-split is automatically a whole number

大提示：

$3 : 2 : 1$ 的条件给出两个线性方程，化简为 $9y = 13z$ ，然后 $x$ 也随之确定；取最小正整数。

The $3 : 2 : 1$ conditions give two linear equations that reduce to $9y = 13z,$ with $x$ then determined; take the smallest positive integers

解答：

设第一只猴子拿走 $8x$ 个香蕉，留下 $6x$ 个，并给另外两只各 $x$ 个；第二只拿走 $8y$ 个，留下 $2y$ 个，并给另外两只各 $3y$ 个；第三只拿走 $24z$ 个，留下 $2z$ 个，并给另外两只各 $11z$ 个。所有分配恰好都是整数，当且仅当 $x$ 、 $y$ 、 $z$ 是正整数。最后三只猴子的数量分别为 $6x + 3y + 11z$ 、 $x + 2y + 11z$ 和 $x + 3y + 2z$ 。

比值 $3 : 2 : 1$ 表示第一份是第三份的三倍，第二份是第三份的两倍： $\begin{aligned} &6x + 3y + 11z = 3(x + 3y + 2z) \\ &\quad \Longrightarrow\quad 3x + 5z = 6y \end{aligned}\text{，}$ $\begin{aligned} &x + 2y + 11z = 2(x + 3y + 2z) \\ &\quad \Longrightarrow\quad x + 4y = 7z \end{aligned}\text{。}$ 将 $x = 7z - 4y$ 代入第一式，得 $26z = 18y$ ，所以 $9y = 13z$ 。因此 $y = 13n$ ， $z = 9n$ ，其中 $n$ 为正整数，进而 $x = 63n - 52n = 11n$ 。

总数为 $8x + 8y + 24z$ $= (88 + 104 + 216)n$ $= 408n$ ，当 $n = 1$ 时最小，答案为 $408$ 。

Say the first monkey takes $8x$ bananas, keeping $6x$ and giving $x$ to each of the others; the second takes $8y,$ keeping $2y$ and giving $3y$ to each; the third takes $24z,$ keeping $2z$ and giving $11z$ to each. All divisions are whole numbers exactly when $x,$ $y,$ $z$ are positive integers. The final amounts are $6x + 3y + 11z,$ $x + 2y + 11z,$ and $x + 3y + 2z.$

The ratio $3 : 2 : 1$ says the first amount is triple the third and the second is double the third: $\begin{aligned} &6x + 3y + 11z = 3(x + 3y + 2z) \\ &\quad \Longrightarrow\quad 3x + 5z = 6y, \end{aligned}$ $\begin{aligned} &x + 2y + 11z = 2(x + 3y + 2z) \\ &\quad \Longrightarrow\quad x + 4y = 7z. \end{aligned}$ Substituting $x = 7z - 4y$ into the first equation gives $26z = 18y,$ so $9y = 13z.$ Thus $y = 13n$ and $z = 9n$ for a positive integer $n,$ and then $x = 63n - 52n = 11n.$

The total is $8x + 8y + 24z$ $= (88 + 104 + 216)n$ $= 408n,$ least when $n = 1:$ the answer is $408.$

7.

$ABCD$ 是一张长方形纸片，折叠后使角 $B$ 与点 $B'$ 重合，而该点位于边 $\overline{AD}$ 上。折痕为 $\overline{EF}$ ，其中 $E$ 在 $\overline{AB}$ 上， $F$ 在 $\overline{CD}$ 上。已知 $AE = 8$ 、 $BE = 17$ 、 $CF = 3$ 。长方形 $ABCD$ 的周长为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

$ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $\overline{AD}.$ The crease is $\overline{EF},$ where $E$ is on $\overline{AB}$ and $F$ is on $\overline{CD}.$ The dimensions $AE = 8,$ $BE = 17,$ and $CF = 3$ are given. The perimeter of rectangle $ABCD$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：293

知识点：折纸勾股定理坐标几何

难度评级：2650

小提示：

折叠保持距离，所以 $B'E = BE = 17$ ；直角三角形 $AEB'$ 给出 $AB' = 15$ ，且 $AB = 25$ 。

Folding preserves distances, so $B'E = BE = 17;$ right triangle $AEB'$ gives $AB' = 15$ and $AB = 25$

大提示：

折痕上的每个点到 $B$ 和 $B'$ 的距离相等，所以 $\overline{EF} \perp \overline{BB'}$ ；用 $EF$ 的斜率确定 $F$ 在 $\overline{CD}$ 上的位置。

Every point of the crease is equidistant from $B$ and $B',$ so $\overline{EF} \perp \overline{BB'};$ use the slope of $EF$ to locate $F$ on $\overline{CD}$

解答：

折叠把 $B$ 关于折痕反射到 $B'$ ，所以 $B'E = BE = 17$ 。在直角三角形 $AEB'$ 中， $AB' = \sqrt{17^2 - 8^2} = 15$ ，且 $AB = AE + EB = 25$ 。取 $A = (0, 0)$ 、 $B = (25, 0)$ 、 $B' = (0, 15)$ 。

折痕上的点到 $B$ 和 $B'$ 等距，所以 $\overline{EF}$ 垂直于 $\overline{BB'}$ 。由于 $BB'$ 的斜率为 $-\frac{3}{5}$ ，过 $E = (8, 0)$ 的折痕斜率为 $\frac{5}{3}$ ，它与直线 $CD$ （高度 $h = BC$ ）相交处的横坐标为 $x = 8 + \frac{3h}{5}$ 。条件 $CF = 3$ 给出 $25 - \left(8 + \frac{3h}{5}\right) = 3\text{，}$ 所以 $h = \frac{70}{3}\text{。}$

周长为 $2\left(25 + \frac{70}{3}\right) = \frac{290}{3}$ ，所以 $m + n = 290 + 3 = 293$ 。

Folding reflects $B$ to $B'$ across the crease, so $B'E = BE = 17.$ In right triangle $AEB',$ $AB' = \sqrt{17^2 - 8^2} = 15,$ and $AB = AE + EB = 25.$ Place $A = (0, 0),$ $B = (25, 0),$ $B' = (0, 15).$

Points on the crease are equidistant from $B$ and $B',$ so $\overline{EF}$ is perpendicular to $\overline{BB'}.$ Since $BB'$ has slope $-\frac{3}{5},$ the crease through $E = (8, 0)$ has slope $\frac{5}{3},$ and it meets the line $CD$ (at height $h = BC$ ) at $x = 8 + \frac{3h}{5}.$ The condition $CF = 3$ gives $25 - \left(8 + \frac{3h}{5}\right) = 3,$ so $h = \frac{70}{3}.$

The perimeter is $2\left(25 + \frac{70}{3}\right) = \frac{290}{3},$ so $m + n = 290 + 3 = 293.$

8.

$2004^{2004}$ 的正整数因子中，有多少个恰好有 $2004$ 个正因数？

How many positive integer divisors of $2004^{2004}$ are divisible by exactly $2004$ positive integers?

答案

答案：54

知识点：因数个数质因数分解隔板法

难度评级：2450

小提示：

因为 $2004 = 2^2 \cdot 3 \cdot 167$ ，所以 $2004^{2004}$ 的一个形如 $2^i 3^j 167^k$ 的因子恰有 $(i+1)(j+1)(k+1)$ 个正因子。

Since $2004 = 2^2 \cdot 3 \cdot 167,$ a divisor $2^i 3^j 167^k$ of $2004^{2004}$ has exactly $(i+1)(j+1)(k+1)$ divisors

大提示：

逐个质因子计数乘积为 $2004$ 的有序三元组：把 $2^2$ 分到三个因子中有 $\binom{4}{2}$ 种，而 $3$ 和 $167$ 各有 $3$ 种去处。

Count ordered triples with product $2004$ prime by prime: split $2^2$ among three factors in $\binom{4}{2}$ ways, and each of $3$ and $167$ in $3$ ways

解答：

因为 $2004 = 2^2 \cdot 3 \cdot 167$ ，所以 $2004^{2004} = 2^{4008} \cdot 3^{2004} \cdot 167^{2004}$ 。它的因子为 $N = 2^i 3^j 167^k$ ，其中 $i \le 4008$ 、 $j, k \le 2004$ 。这样的 $N$ 有 $(i+1)(j+1)(k+1)$ 个因子，所以需要 $(i+1)(j+1)(k+1) = 2004$ 。

每个乘积为 $2004$ 的正整数有序三元组都会给出可行指数，因为每个因子都不超过 $2004$ 。逐个质因子计数：质数 $2$ 的指数 $2$ 分到三个因子中，由隔板法有 $\binom{2+2}{2} = 6$ 种；质数 $3$ 和 $167$ 各自都有 $3$ 个因子可供选择。

总数为 $6 \cdot 3 \cdot 3 = 54$ 。

Since $2004 = 2^2 \cdot 3 \cdot 167,$ we have $2004^{2004} = 2^{4008} \cdot 3^{2004} \cdot 167^{2004},$ so its divisors are $N = 2^i 3^j 167^k$ with $i \le 4008$ and $j, k \le 2004.$ Such an $N$ has $(i+1)(j+1)(k+1)$ divisors, so we need $(i+1)(j+1)(k+1) = 2004.$

Every ordered triple of positive integers with product $2004$ yields admissible exponents, since each factor is at most $2004.$ Counting prime by prime: the exponent $2$ of the prime $2$ is split among the three factors in $\binom{2+2}{2} = 6$ ways by stars and bars, and each of the primes $3$ and $167$ goes to one of the $3$ factors.

The count is $6 \cdot 3 \cdot 3 = 54.$

9.

一个正整数数列满足 $a_1 = 1$ 且 $a_9 + a_{10} = 646$ ，其构造方式为：前三项成等比数列，第二、三、四项成等差数列；一般地，对所有 $n \ge 1$ ， $a_{2n-1}$ 、 $a_{2n}$ 、 $a_{2n+1}$ 成等比数列，而 $a_{2n}$ 、 $a_{2n+1}$ 、 $a_{2n+2}$ 成等差数列。令 $a_n$ 为该数列中小于 $1000$ 的最大项。求 $n + a_n$ 。

A sequence of positive integers with $a_1 = 1$ and $a_9 + a_{10} = 646$ is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all $n \ge 1,$ the terms $a_{2n-1},$ $a_{2n},$ and $a_{2n+1}$ are in geometric progression, and the terms $a_{2n},$ $a_{2n+1},$ and $a_{2n+2}$ are in arithmetic progression. Let $a_n$ be the greatest term in this sequence that is less than $1000.$ Find $n + a_n.$

答案

答案：973

知识点：等比数列等差数列找规律

难度评级：2840

小提示：

令 $a_2 = r$ 。每个等比或等差条件都会确定下一项，从而得到 $a_9 = (4r-3)^2$ 和 $a_{10} = (4r-3)(5r-4)$ 。

Let $a_2 = r;$ each progression condition forces the next term, giving $a_9 = (4r-3)^2$ and $a_{10} = (4r-3)(5r-4)$

大提示：

因而 $(4r-3)(9r-7) = 646$ ，推出 $r = 5$ ；奇数下标项为 $(2n-1)^2$ ，偶数下标项为 $(2n-3)(2n+1)$ 。

Then $(4r-3)(9r-7) = 646$ forces $r = 5,$ so odd-indexed terms are $(2n-1)^2$ and even-indexed terms are $(2n-3)(2n+1)$

解答：

令 $a_2 = r$ 。等比条件给出 $a_3 = r^2$ ，等差条件给出 $a_4 = 2r^2 - r = r(2r-1)$ ，接着 $a_5 = (2r-1)^2$ ，依此类推可归纳得到 $\begin{aligned} a_{2k+1} &= \bigl(kr - (k-1)\bigr)^2, \\ a_{2k+2} &= \bigl(kr - (k-1)\bigr) \\ &\quad {}\cdot \bigl((k+1)r - k\bigr) \end{aligned}\text{。}$ 特别地， $a_9 = (4r-3)^2$ ， $a_{10} = (4r-3)(5r-4)$ ，所以 $a_9 + a_{10} = (4r-3)(9r-7)$ $= 646$ 。展开得 $36r^2 - 55r - 625 = 0$ ，分解为 $(r - 5)(36r + 125) = 0$ ，所以 $r = 5$ 。

当 $r = 5$ 时， $kr - (k-1) = 4k + 1$ ，所以 $a_{2k+1} = (4k+1)^2$ ， $a_{2k+2} = (4k+1)(4k+5)$ ，数列递增。因为 $a_{17} = 33^2 = 1089 \gt 1000$ ，而 $a_{16} = 29 \cdot 33 = 957$ ，所以小于 $1000$ 的最大项是 $a_{16} = 957$ 。

因此 $n + a_n = 16 + 957 = 973$ 。

Let $a_2 = r.$ The geometric condition gives $a_3 = r^2,$ the arithmetic condition gives $a_4 = 2r^2 - r = r(2r-1),$ then $a_5 = (2r-1)^2,$ and so on: inductively $\begin{aligned} a_{2k+1} &= \bigl(kr - (k-1)\bigr)^2, \\ a_{2k+2} &= \bigl(kr - (k-1)\bigr) \\ &\quad {}\cdot \bigl((k+1)r - k\bigr). \end{aligned}$ In particular $a_9 = (4r-3)^2$ and $a_{10} = (4r-3)(5r-4),$ so $a_9 + a_{10} = (4r-3)(9r-7)$ $= 646.$ Expanding gives $36r^2 - 55r - 625 = 0,$ which factors as $(r - 5)(36r + 125) = 0,$ so $r = 5.$

With $r = 5$ we get $kr - (k-1) = 4k + 1,$ so $a_{2k+1} = (4k+1)^2$ and $a_{2k+2} = (4k+1)(4k+5);$ the sequence is increasing. Since $a_{17} = 33^2 = 1089 \gt 1000$ while $a_{16} = 29 \cdot 33 = 957,$ the greatest term below $1000$ is $a_{16} = 957.$

Therefore $n + a_n = 16 + 957 = 973.$

10.

设 $\mathcal{S}$ 是 $1$ 到 $2^{40}$ 之间、二进制表示中恰有两个 $1$ 的整数集合。从 $\mathcal{S}$ 中随机选一个数，它能被 $9$ 整除的概率为 $\frac{p}{q}$ ，其中 $p$ 和 $q$ 是互质正整数。求 $p + q$ 。

Let $\mathcal{S}$ be the set of integers between $1$ and $2^{40}$ whose binary expansions have exactly two $1$ ’s. If a number is chosen at random from $\mathcal{S},$ the probability that it is divisible by $9$ is $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

答案

答案：913

知识点：进制模幂运算乘法阶数对计数

难度评级：2920

小提示：

$\mathcal{S}$ 中的元素为 $2^a + 2^b = 2^a(2^{b-a} + 1)$ ，其中 $a \lt b$ ，所以能否被 $9$ 整除只取决于 $b - a$ 。

Elements of $\mathcal{S}$ are $2^a + 2^b = 2^a(2^{b-a} + 1)$ with $a \lt b,$ so divisibility by $9$ depends only on $b - a$

大提示：

$2$ 的幂模 $9$ 的周期为 $6$ ，且 $2^d \equiv 8$ 当且仅当 $d \equiv 3 \pmod{6}$ ；数出每个这种差值对应的数对。

Powers of $2$ repeat mod $9$ with period $6,$ and $2^d \equiv 8$ exactly when $d \equiv 3 \pmod{6};$ count the pairs with each such difference

解答：

集合 $\mathcal{S}$ 由 $\binom{40}{2} = 780$ 个数 $2^a + 2^b$ 组成，其中 $0 \le a \lt b \le 39$ 。由于 $2^a$ 与 $9$ 互质， $2^a(2^{b-a} + 1)$ 能被 $9$ 整除当且仅当 $2^{b-a} \equiv -1 \pmod{9}$ 。 $2$ 的幂模 $9$ 依次循环为 $2, 4, 8, 7, 5, 1$ ，周期为 $6$ ，所以 $2^d \equiv 8 \equiv -1$ 当且仅当 $d \equiv 3 \pmod{6}$ 。

对每个差值 $d = b - a$ 有 $40 - d$ 个数对，所以 $9$ 的倍数在 $\mathcal{S}$ 中的个数为 $\begin{aligned} &\sum_{d = 3, 9, \ldots, 39} (40 - d) \\ &= 37 + 31 + 25 \\ &\quad {}+ 19 + 13 + 7 + 1 \\ &= 133 \end{aligned}\text{。}$

概率为 $\frac{133}{780}$ 。因为 $133 = 7 \cdot 19$ ，而 $780 = 2^2 \cdot 3 \cdot 5 \cdot 13$ ，该分数已最简。因此 $p + q = 133 + 780 = 913$ 。

The set $\mathcal{S}$ consists of the $\binom{40}{2} = 780$ numbers $2^a + 2^b$ with $0 \le a \lt b \le 39.$ Since $2^a$ is coprime to $9,$ we have $2^a(2^{b-a} + 1)$ divisible by $9$ exactly when $2^{b-a} \equiv -1 \pmod{9}.$ The powers of $2$ modulo $9$ cycle through $2, 4, 8, 7, 5, 1$ with period $6,$ so $2^d \equiv 8 \equiv -1$ exactly when $d \equiv 3 \pmod{6}.$

For each difference $d = b - a$ there are $40 - d$ pairs, so the number of multiples of $9$ in $\mathcal{S}$ is $\begin{aligned} &\sum_{d = 3, 9, \ldots, 39} (40 - d) \\ &= 37 + 31 + 25 \\ &\quad {}+ 19 + 13 + 7 + 1 \\ &= 133. \end{aligned}$

The probability is $\frac{133}{780},$ and since $133 = 7 \cdot 19$ while $780 = 2^2 \cdot 3 \cdot 5 \cdot 13,$ it is in lowest terms. Thus $p + q = 133 + 780 = 913.$

11.

一个直圆锥的底面半径为 $600$ ，高为 $200\sqrt{7}$ 。一只苍蝇从圆锥表面上一点出发，该点到圆锥顶点的距离为 $125$ ，并沿圆锥表面爬到圆锥正对面的一个点，该点到顶点的距离为 $375\sqrt{2}$ 。求苍蝇可能爬行的最短距离。

A right circular cone has a base with radius $600$ and height $200\sqrt{7}.$ A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is $125,$ and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is $375\sqrt{2}.$ Find the least distance that the fly could have crawled.

答案

答案：625

知识点：圆锥展开图（立体几何）余弦定理

难度评级：2990

小提示：

斜高为 $\sqrt{600^2 + (200\sqrt{7})^2} = 800$ ；将圆锥展开成半径为 $800$ 的扇形。

The slant height is $\sqrt{600^2 + (200\sqrt{7})^2} = 800;$ unroll the cone into a circular sector of radius $800$

大提示：

扇形圆心角为 $270^\circ$ ，所以正对面的点在扇形中相隔 $135^\circ$ ；使用余弦定理。

The sector’s central angle is $270^\circ,$ so exactly opposite points are $135^\circ$ apart in the sector; apply the law of cosines

解答：

圆锥斜高为 $\sqrt{600^2 + (200\sqrt{7})^2}$ $= \sqrt{360000 + 280000}$ $= 800$ 。沿经过起点的母线剪开并展开，得到半径为 $800$ 的扇形，其弧长等于底面周长 $2\pi \cdot 600 = 1200\pi$ 。半径为 $800$ 的整圆周长为 $1600\pi$ ，所以扇形的圆心角为 $\frac{3}{4} \cdot 360^\circ = 270^\circ$ 。圆锥正对面的点相当于绕底面半圈，在展开扇形中相隔 $135^\circ$ 。

最短路径是展开图中两个点之间的线段，这两个点到顶点的距离分别为 $125$ 和 $375\sqrt{2}$ ，夹角为 $135^\circ$ 。由余弦定理， $\begin{aligned} d^2 &= 125^2 + (375\sqrt{2})^2 \\ &\quad {}- 2 \cdot 125 \cdot 375\sqrt{2} \cos 135^\circ \\ &= 15625 + 281250 + 93750 \\ &= 390625 \end{aligned}\text{。}$

因此 $d = 625$ 。

The slant height is $\sqrt{600^2 + (200\sqrt{7})^2}$ $= \sqrt{360000 + 280000}$ $= 800.$ Cutting the cone along the ruling through the starting point and unrolling gives a sector of radius $800$ whose arc has the base circumference $2\pi \cdot 600 = 1200\pi;$ since a full circle of radius $800$ has circumference $1600\pi,$ the central angle is $\frac{3}{4} \cdot 360^\circ = 270^\circ.$ A point on the exact opposite side of the cone is halfway around, which in the unrolled sector is $135^\circ$ away.

The shortest crawl is the straight segment between the two points, at radii $125$ and $375\sqrt{2}$ with a $135^\circ$ angle between them. By the law of cosines, $\begin{aligned} d^2 &= 125^2 + (375\sqrt{2})^2 \\ &\quad {}- 2 \cdot 125 \cdot 375\sqrt{2} \cos 135^\circ \\ &= 15625 + 281250 + 93750 \\ &= 390625. \end{aligned}$

Thus $d = 625.$

12.

设 $ABCD$ 是等腰梯形，其边长为 $AB = 6$ 、 $BC = 5 = DA$ 、 $CD = 4$ 。作半径为 $3$ 、圆心分别为 $A$ 和 $B$ 的两个圆，再作半径为 $2$ 、圆心分别为 $C$ 和 $D$ 的两个圆。有一个位于梯形内部的圆与这四个圆都相切。它的半径为 $\frac{-k + m\sqrt{n}}{p}$ ，其中 $k$ 、 $m$ 、 $n$ 、 $p$ 是正整数， $n$ 不被任何质数的平方整除，且 $k$ 与 $p$ 互质。求 $k + m + n + p$ 。

Let $ABCD$ be an isosceles trapezoid, whose dimensions are $AB = 6,$ $BC = 5 = DA,$ and $CD = 4.$ Draw circles of radius $3$ centered at $A$ and $B,$ and circles of radius $2$ centered at $C$ and $D.$ A circle contained within the trapezoid is tangent to all four of these circles. Its radius is $\frac{-k + m\sqrt{n}}{p},$ where $k,$ $m,$ $n,$ and $p$ are positive integers, $n$ is not divisible by the square of any prime, and $k$ and $p$ are relatively prime. Find $k + m + n + p.$

答案

答案：134

知识点：相切圆梯形勾股定理根式

难度评级：3060

小提示：

梯形的高为 $\sqrt{24}$ 。内圆圆心在对称轴上；相切条件表明，圆心到 $A$ 、 $B$ 的距离均为 $x + 3$ ，到 $C$ 、 $D$ 的距离均为 $x + 2$ 。

The trapezoid’s height is $\sqrt{24}.$ Center the inner circle on the axis of symmetry; tangency means its center is at distance $x + 3$ from $A$ and $B,$ and $x + 2$ from $C$ and $D$

大提示：

距离 $\sqrt{x^2 + 6x}$ 表示圆心高出 $\overline{AB}$ 的高度，而距离 $\sqrt{x^2 + 4x}$ 表示圆心低于 $\overline{CD}$ 的高度；两者之和为 $\sqrt{24}$ 。

The center lies $\sqrt{x^2 + 6x}$ above $\overline{AB}$ and $\sqrt{x^2 + 4x}$ below $\overline{CD},$ and those two heights add to $\sqrt{24}$

解答：

从 $C$ 和 $D$ 作垂线可知，每条长为 $5$ 的腰对应的水平偏移为 $\frac{6 - 4}{2} = 1$ ，所以梯形高度为 $\sqrt{25 - 1} = \sqrt{24}$ 。由对称性，内圆圆心 $O$ 位于竖直对称轴上；该轴经过点 $E$ ，即 $\overline{AB}$ 的中点，也经过点 $F$ ，即 $\overline{CD}$ 的中点。若内圆半径为 $x$ ，外切条件给出 $OA = x + 3$ 和 $OC = x + 2$ 。又因为 $AE = 3$ 、 $CF = 2$ ，所以 $\begin{aligned} OE &= \sqrt{(x+3)^2 - 9} \\ &= \sqrt{x^2 + 6x}, \\ OF &= \sqrt{(x+2)^2 - 4} \\ &= \sqrt{x^2 + 4x} \end{aligned}\text{。}$

因为 $OE + OF = \sqrt{24}$ ，移项后平方得 $\sqrt{24(x^2 + 4x)} = 12 - x$ ，再次平方得到 $24x^2 + 96x = 144 - 24x + x^2$ ，即 $23x^2 + 120x - 144 = 0$ 。

正根为 $\begin{aligned} x &= \frac{-120 + \sqrt{14400 + 13248}}{46} \\ &= \frac{-120 + 96\sqrt{3}}{46} \\ &= \frac{-60 + 48\sqrt{3}}{23} \end{aligned}\text{，}$ 所以 $k + m + n + p$ $= 60 + 48 + 3 + 23$ $= 134$ 。

Dropping perpendiculars from $C$ and $D$ shows each leg of length $5$ spans a horizontal offset of $\frac{6 - 4}{2} = 1,$ so the height of the trapezoid is $\sqrt{25 - 1} = \sqrt{24}.$ By symmetry the inner circle’s center $O$ lies on the vertical axis through the midpoints $E$ of $\overline{AB}$ and $F$ of $\overline{CD}.$ If its radius is $x,$ external tangency gives $OA = x + 3$ and $OC = x + 2,$ so with $AE = 3$ and $CF = 2,$ $\begin{aligned} OE &= \sqrt{(x+3)^2 - 9} \\ &= \sqrt{x^2 + 6x}, \\ OF &= \sqrt{(x+2)^2 - 4} \\ &= \sqrt{x^2 + 4x}. \end{aligned}$

Since $OE + OF = \sqrt{24},$ moving one radical across and squaring gives $\sqrt{24(x^2 + 4x)} = 12 - x,$ and squaring again yields $24x^2 + 96x = 144 - 24x + x^2,$ that is $23x^2 + 120x - 144 = 0.$

The positive root is $\begin{aligned} x &= \frac{-120 + \sqrt{14400 + 13248}}{46} \\ &= \frac{-120 + 96\sqrt{3}}{46} \\ &= \frac{-60 + 48\sqrt{3}}{23}, \end{aligned}$ so $k + m + n + p$ $= 60 + 48 + 3 + 23$ $= 134.$

13.

设 $ABCDE$ 是一个凸五边形，且 $\overline{AB} \parallel \overline{CE}$ ， $\overline{BC} \parallel \overline{AD}$ ， $\overline{AC} \parallel \overline{DE}$ ， $\angle ABC = 120^\circ$ ， $AB = 3$ ， $BC = 5$ ， $DE = 15$ 。已知三角形 $ABC$ 的面积与三角形 $EBD$ 的面积之比为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

Let $ABCDE$ be a convex pentagon with $\overline{AB} \parallel \overline{CE},$ $\overline{BC} \parallel \overline{AD},$ $\overline{AC} \parallel \overline{DE},$ $\angle ABC = 120^\circ,$ $AB = 3,$ $BC = 5,$ and $DE = 15.$ Given that the ratio between the area of triangle $ABC$ and the area of triangle $EBD$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

答案

答案：484

知识点：相似平行四边形余弦定理面积比

难度评级：3160

小提示：

在三角形 $ABC$ 中用余弦定理得 $AC = 7$ 。令 $F$ 为 $\overline{AD}$ 与 $\overline{CE}$ 的交点，则 $ABCF$ 是平行四边形。

The law of cosines in triangle $ABC$ gives $AC = 7.$ Let $F$ be the intersection of $\overline{AD}$ and $\overline{CE};$ then $ABCF$ is a parallelogram

大提示：

三角形 $FAC$ 与 $FDE$ 相似，比例为 $7 : 15$ 。令 $h$ 表示点 $B$ 到 $\overline{AC}$ 的距离，再用它表示各点到平行线 $AC$ 和 $DE$ 的距离。

Triangles $FAC$ and $FDE$ are similar in ratio $7 : 15,$ so measure every distance to the parallel lines $AC$ and $DE$ in terms of the distance $h$ from $B$ to $\overline{AC}$

解答：

由余弦定理， $AC^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cos 120^\circ$ $= 49$ ，所以 $AC = 7$ 。令 $F$ 为 $\overline{AD}$ 与 $\overline{CE}$ 的交点。因为 $AF \parallel BC$ 且 $CF \parallel AB$ ，四边形 $ABCF$ 是平行四边形，所以 $F$ 和 $B$ 到直线 $AC$ 的距离均为 $h$ ，且位于该直线两侧。于是 $[ABC] = \frac{1}{2} \cdot 7h$ 。

因为 $\overline{AC} \parallel \overline{DE}$ ，三角形 $FAC$ 与 $FDE$ 相似，比例为 $AC : DE = 7 : 15$ ，所以 $F$ 到直线 $DE$ 的距离为 $\frac{15h}{7}$ ，且 $DE$ 在 $F$ 远离 $AC$ 的一侧。因此 $B$ 到直线 $DE$ 的距离为 $h + h + \frac{15h}{7} = \frac{29h}{7}$ ，从而 $[EBD] = \frac{1}{2} \cdot 15 \cdot \frac{29h}{7} = \frac{435h}{14}\text{。}$

因此 $\frac{[ABC]}{[EBD]} = \frac{\frac{7h}{2}}{\frac{435h}{14}} = \frac{49}{435}$ 。由于 $435 = 3 \cdot 5 \cdot 29$ ，该分数已最简，答案为 $m + n = 49 + 435 = 484$ 。

By the law of cosines, $AC^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cos 120^\circ$ $= 49,$ so $AC = 7.$ Let $F$ be the intersection of $\overline{AD}$ and $\overline{CE}.$ Since $AF \parallel BC$ and $CF \parallel AB,$ quadrilateral $ABCF$ is a parallelogram, so $F$ lies at the same distance $h$ from line $AC$ as $B,$ on the opposite side, where $[ABC] = \frac{1}{2} \cdot 7h.$

Since $\overline{AC} \parallel \overline{DE},$ triangles $FAC$ and $FDE$ are similar with ratio $AC : DE = 7 : 15,$ so the distance from $F$ to line $DE$ is $\frac{15h}{7},$ with $DE$ on the far side of $F$ from $AC.$ The distance from $B$ to line $DE$ is therefore $h + h + \frac{15h}{7} = \frac{29h}{7},$ giving $[EBD] = \frac{1}{2} \cdot 15 \cdot \frac{29h}{7} = \frac{435h}{14}.$

Thus $\frac{[ABC]}{[EBD]} = \frac{\frac{7h}{2}}{\frac{435h}{14}} = \frac{49}{435},$ which is in lowest terms since $435 = 3 \cdot 5 \cdot 29.$ The answer is $m + n = 49 + 435 = 484.$

14.

考虑一个由 $n$ 个 $7$ 组成的字符串 $7777\ldots77$ ，在其中插入 $+$ 号可以得到一个算式。例如，算式 $7 + 77 + 777 + 7 + 7 = 875$ 可以由八个 $7$ 这样得到。有多少个 $n$ 的取值，能通过插入 $+$ 号使所得算式的值为 $7000$ ？

Consider a string of $n$ $7$ ’s, $7777\ldots77,$ into which $+$ signs are inserted to produce an arithmetic expression. For example, $7 + 77 + 777 + 7 + 7 = 875$ could be obtained from eight $7$ ’s in this way. For how many values of $n$ is it possible to insert $+$ signs so that the resulting expression has value $7000?$

答案

答案：108

知识点：丢番图方程数字分类讨论

难度评级：3270

小提示：

全部除以 $7$ ：需要 $x$ 个 $1$ 、 $y$ 个 $11$ 、 $z$ 个 $111$ ，满足 $x + 11y + 111z = 1000$ ，且 $n = x + 2y + 3z$ 。

Divide everything by $7:$ you need $x$ copies of $1,$ $y$ of $11,$ and $z$ of $111$ with $x + 11y + 111z = 1000$ and $n = x + 2y + 3z$

大提示：

相减得 $n = 1000 - 9(y + 12z)$ ；确定 $y + 12z$ 在条件 $11y + 111z \le 1000$ 下恰能取到哪些值。

Subtracting shows $n = 1000 - 9(y + 12z);$ determine exactly which values $y + 12z$ can reach when $11y + 111z \le 1000$

解答：

除以 $7$ 后，各项只能是 $1$ 、 $11$ 或 $111$ （再长的 $1111 \gt 1000$ ），而目标变为 $1000$ 。设 $x$ 、 $y$ 、 $z$ 分别为这三种项的个数，则 $x + 11y + 111z = 1000$ ，且 $n = x + 2y + 3z$ 。相减得 $n = 1000 - 9(y + 12z)\text{，}$ 所以可能的 $n$ 与可达到的 $v = y + 12z$ 值一一对应。

约束为 $z \le 9$ ，且 $0 \le y \le \frac{1000 - 111z}{11}$ （这样 $x \ge 0$ ）。当 $z = 0, 1, \ldots, 9$ 时， $v = y + 12z$ 的取值区间分别为 $[0, 90]$ 、 $[12, 92]$ 、 $[24, 94]$ 、 $[36, 96]$ 、 $[48, 98]$ 、 $[60, 100]$ 、 $[72, 102]$ 、 $[84, 104]$ 、 $[96, 106]$ ，以及 $\{108\}$ （当 $z = 9$ 时只有 $y = 0$ 可行）。它们的并集是从 $0$ 到 $106$ 的所有整数，再加上 $108$ ；只有 $107$ 不可达到。

因此 $v$ 有 $107 + 1 = 108$ 个取值， $n = 1000 - 9v$ 也有 $108$ 个取值。

Dividing by $7$ turns the summands into $1,$ $11,$ or $111$ (no summand with four or more digits fits, since $1111 \gt 1000$ ) and the target into $1000.$ If $x,$ $y,$ $z$ count the summands of each size, then $x + 11y + 111z = 1000$ and $n = x + 2y + 3z.$ Subtracting gives $n = 1000 - 9(y + 12z),$ so the possible $n$ correspond exactly to the attainable values of $v = y + 12z.$

The constraints are $z \le 9$ and $0 \le y \le \frac{1000 - 111z}{11}$ (then $x \ge 0$ follows). For $z = 0, 1, \ldots, 9$ the value $v = y + 12z$ ranges over the intervals $[0, 90],$ $[12, 92],$ $[24, 94],$ $[36, 96],$ $[48, 98],$ $[60, 100],$ $[72, 102],$ $[84, 104],$ $[96, 106],$ and $\{108\}$ (when $z = 9,$ only $y = 0$ fits). Their union is every integer from $0$ to $106$ together with $108;$ only $107$ is unattainable.

So $v$ takes $107 + 1 = 108$ values, and $n = 1000 - 9v$ takes $108$ values.

15.

一条细长纸带长 $1024$ 个单位、宽 $1$ 个单位，并被分成 $1024$ 个单位正方形。纸带反复对折。第一次折叠时，将纸带右端折到与左端重合并叠在其上，得到一条 $512$ 乘 $1$ 、双层厚的纸带。接着，再把这条纸带的右端折到与左端重合并叠在其上，得到一条 $256$ 乘 $1$ 、四层厚的纸带。这个过程再重复 $8$ 次。最后一次折叠后，纸带变成一叠 $1024$ 个单位正方形。原来从左数第 $942$ 个正方形的下面有多少个正方形？

A long thin strip of paper is $1024$ units in length, $1$ unit in width, and is divided into $1024$ unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a $512$ by $1$ strip of double thickness. Next, the right end of this strip is folded over to coincide with and lie on top of the left end, resulting in a $256$ by $1$ strip of quadruple thickness. This process is repeated $8$ more times. After the last fold, the strip has become a stack of $1024$ unit squares. How many of these squares lie below the square that was originally the $942$ nd square counting from the left?

答案

答案：593

知识点：折纸过程模拟不变量

难度评级：3500

小提示：

同时追踪该正方形从左端数的位置和从底部数的位置；对左半部分的正方形，一次折叠保持这两个位置不变。

Track the square’s position counted from the left end and from the bottom of the stack; a fold leaves both unchanged for squares in the left half

大提示：

对右半部分的正方形，新位置从左数等于旧位置从右数，新位置从上数等于旧位置从底部数。

For a square in the right half, the new position from the left is the old position from the right, and the new position from the top is the old position from the bottom

解答：

经过 $f$ 次折叠后，纸带长 $2^{10-f}$ 个正方形，厚 $2^f$ 层。因此从左数的位置 $L$ 与从右数的位置 $R$ 满足 $L + R = 2^{10-f} + 1$ ，从底部数的位置 $B$ 与从顶部数的位置 $T$ 满足 $B + T = 2^f + 1$ 。右半部分折到左半部分时，左半部分的正方形保持 $L$ 和 $B$ 不变；右半部分的正方形被翻转，新的 $L$ 是旧的 $R$ ，新的 $T$ 是旧的 $B$ 。

第 $942$ 个正方形起始为 $(L, B) = (942, 1)$ 。逐次应用规则，十次折叠后的状态为 $\begin{gathered} (83, 2),\ (83, 2),\ (83, 2),\ (46, 15),\\ \ (19, 18),\ (14, 47),\ (3, 82),\\ \ (3, 82),\ (2, 431),\ (1, 594) \end{gathered}\text{。}$ 例如第四次折叠时纸带长度为 $128$ ，且 $L = 83 \gt 64$ ，所以新的 $L$ 为 $128 + 1 - 83 = 46$ ，新的 $T$ 是旧的 $B = 2$ ，从而 $B = 16 + 1 - 2 = 15$ 。

在最后的 $1024$ 层堆叠中，该正方形从底部数位于第 $594$ 层，所以其下方有 $594 - 1 = 593$ 个正方形。

After $f$ folds the strip is $2^{10-f}$ squares long and $2^f$ layers thick, so the positions $L$ from the left and $R$ from the right satisfy $L + R = 2^{10-f} + 1,$ and the positions $B$ from the bottom and $T$ from the top satisfy $B + T = 2^f + 1.$ When the right half is folded over onto the left, a square in the left half keeps its $L$ and $B,$ while a square in the right half is flipped: its new $L$ is its old $R,$ and its new $T$ is its old $B.$

The $942$ nd square starts at $(L, B) = (942, 1).$ Applying the rule through the ten folds gives $\begin{gathered} (83, 2),\ (83, 2),\ (83, 2),\ (46, 15),\\ \ (19, 18),\ (14, 47),\ (3, 82),\\ \ (3, 82),\ (2, 431),\ (1, 594). \end{gathered}$ For example, at the fourth fold the strip has length $128$ and $L = 83 \gt 64,$ so the new $L$ is $128 + 1 - 83 = 46$ and the new $T$ is the old $B = 2,$ making $B = 16 + 1 - 2 = 15.$

In the final stack of $1024$ squares, this square sits at height $594$ from the bottom, so $594 - 1 = 593$ squares lie below it.