2008 AIME I 真题
计时
3:00:00
1.
在参加学校派对的学生中， 是女生， 喜欢跳舞。后来又来了 名男生，他们都喜欢跳舞，此时派对上的学生中女生占 。现在派对上有多少名学生喜欢跳舞？
Of the students attending a school party, of the students are girls, and of the students like to dance. After these students are joined by more boy students, all of whom like to dance, the party is now girls. How many students now at the party like to dance?
小提示：
设原来有 名学生； 名男生到来时，女生人数不变
Let be the original number of students; the number of girls does not change when the boys arrive
大提示：
解 ，再计算原来学生中的 加上新来的 名男生
Solve then count of the original students plus all new boys
解答：
设原来派对上有 名学生，则女生有 人，喜欢跳舞的有 人。 名男生到来后，女生人数不变，总人数变为 ，所以 。于是 ，得 。
现在喜欢跳舞的学生人数为 。
Let be the number of students originally at the party, so are girls and like to dance. When the boys arrive, the number of girls is unchanged but the total becomes so Then giving
The number of students who now like to dance is
2.
正方形 的边长为 个单位。等腰三角形 的底边为 ，并且三角形 与正方形 的公共部分面积为 平方单位。求 所对应的 的高的长度。
Square has sides of length units. Isosceles triangle has base and the area common to triangle and square is square units. Find the length of the altitude to in
小提示：
如果高 不超过 ，重叠部分就是整个三角形，会迫使 ，矛盾；因此顶点在正方形外
If the altitude were at most the overlap would be the whole triangle, forcing — impossible, so the apex pokes out above the square
大提示：
重叠面积等于三角形 的面积减去正方形上方的相似小三角形面积，小三角形的高为 ，底为
The overlap is triangle minus a similar triangle above the square with height and base
解答：
这里 是正方形的一条边。设 为三角形 的高。若 ，三角形会完全在正方形内，其面积 会给出 ，矛盾。所以 ，顶点 在正方形外；对边 截出一个与 相似的小三角形，它的高为 ，底为 。
公共部分是三角形 减去这个小三角形：两边乘以 ，得 ，所以 ，。
Here is a side of the square. Let be the altitude of triangle If the triangle would lie entirely inside the square, and its area would force a contradiction. So and the apex lies outside the square; the opposite side cuts off a smaller triangle similar to with height and base
The common region is triangle minus that small triangle: Multiplying by gives so and
3.
Ed 和 Sue 骑自行车的速度相同且恒定。同样地，他们慢跑的速度相同且恒定，游泳的速度也相同且恒定。Ed 共行进 千米，其中骑自行车 小时、慢跑 小时、游泳 小时；Sue 共行进 千米，其中慢跑 小时、游泳 小时、骑自行车 小时。他们骑自行车、慢跑和游泳的速度都是每小时整数千米。求 Ed 骑自行车、慢跑和游泳速度的平方和。
Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers kilometers after biking for hours, jogging for hours, and swimming for hours, while Sue covers kilometers after jogging for hours, swimming for hours, and biking for hours. Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour. Find the sum of the squares of Ed’s biking, jogging, and swimming rates.
小提示：
两次行程给出整数方程 和
The two trips give and in whole numbers
大提示：
消去 得 ；它的三个正整数解中，只有一个会使 为整数
Eliminate to get of its three positive solutions, only one makes a whole number
解答：
设 、、 分别为骑自行车、慢跑和游泳的速度。两次行程给出 和 将第一个方程乘以两倍后减去第二个方程，得 ，其正整数解为 、 和 。
对应的 的值分别为 、 和 ，所以只有 给出整数速度 。平方和为 。
Let and be the biking, jogging, and swimming rates. The two trips give and Doubling the first equation and subtracting the second yields whose positive integer solutions are and
The corresponding values of are and so only gives a whole-number rate, The sum of the squares is
4.
存在唯一一组正整数 和 满足方程 。求 。
There exist unique positive integers and that satisfy the equation Find
小提示：
将 与 比较
Compare with
大提示：
写成 并分解：，且平方差的两个因子都必须是偶数
Write and factor: and both factors of the difference of squares must be even
解答：
配方得 ，所以 ，因式分解为 这两个因子奇偶性相同，且乘积为偶数，因此二者都是偶数：，。
相加得 ，于是 ；确有 。因此 。
Completing the square, so which factors as The two factors have the same parity, and their product is even, so both are even: and
Adding gives and then indeed Therefore
5.
一个直圆锥的底面半径为 ，高为 。圆锥侧放在平桌上。当圆锥在桌面上无滑动滚动时，圆锥底面与桌面接触的点会描出一段圆弧，其圆心是顶点接触桌面的点。圆锥完成 整圈滚动后第一次回到桌面上的原始位置。 的值可写成 的形式，其中 和 为正整数，且 不被任何质数的平方整除。求 。
A right circular cone has base radius and height The cone lies on its side on a flat table. As the cone rolls on the surface of the table without slipping, the point where the cone’s base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making complete rotations. The value of can be written in the form where and are positive integers and is not divisible by the square of any prime. Find
小提示：
顶点保持不动，所以接触点描出的圆半径是斜高
The vertex stays fixed, so the point of contact traces a circle whose radius is the slant height
大提示：
十七圈意味着 ；将 平方并化简
Seventeen rotations means square and simplify
解答：
底面的接触点与固定顶点的距离始终为 （斜高），所以它描出半径为 的圆。无滑动滚动时，圆锥每滚过一个底面周长的弧长就转一圈，因此恰好 圈后回到原位意味着 即
平方得 ，所以 ，于是 。
The contact point of the base stays at distance (the slant height) from the fixed vertex, so it traces a circle of radius Rolling without slipping, the cone makes one rotation for each base circumference of arc, so returning after exactly rotations means i.e.
Squaring gives so and
6.
一个三角形数阵的第一行由奇整数 、、、、 按递增顺序组成。第一行下面的每一行都比上一行少一个数，最底下一行只有一个数。顶行之后任意一行中的每个数，都等于它正上方一行中斜上方相邻两个数的和。这个数阵中有多少个数是 的倍数？
A triangular array of numbers has a first row consisting of the odd integers in increasing order. Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry. Each entry in any row after the top row equals the sum of the two entries diagonally above it in the row immediately above it. How many entries in the array are multiples of
小提示：
证明编号为 的项在第 行中的值是 ，所以能否被 整除只取决于奇因子
Show the th entry of row is so only the odd factor matters for divisibility by
大提示：
在第 行中，因子 取遍所有与 同奇偶、从 到 的值，所以它唯一可能碰到的 的倍数是 本身
Across row the factor takes every value of ’s parity from to so the only multiple of it can hit is itself
解答：
用归纳法可得，编号为 的项在第 行中的值为 ：第 行给出 ，而第 行相邻两项相加得到 ，这正是第 行的公式。第 行有 项，所以 。
因为 是奇数，一个数是 的倍数当且仅当 能被 整除。当 在第 行中变化时， 取值为 ，它们都与 同奇偶且都小于 。所以唯一可能的 的倍数是 本身，这要求 为奇数且 ，即 。
每个奇数行 都恰有一个这样的数，总数为 。
By induction, the th entry of row is row gives and summing two adjacent entries of row gives the formula for row Row has entries, so
Since is odd, an entry is a multiple of exactly when is divisible by As runs through row the quantity takes the values all with the same parity as and all less than So the only possible multiple of is itself, which requires odd and that is,
Each odd row contains exactly one such entry, for a total of
7.
设 为所有整数 组成的集合，这些整数满足 。例如， 是集合 。在 、、、、 这些集合中，有多少个集合不含完全平方数？
Let be the set of all integers such that For example, is the set How many of the sets do not contain a perfect square?
小提示：
相邻平方数之差为 ；当这个差小于 ，即 时，直到包含 的那一组为止，每组都有平方数
Consecutive squares differ by which is less than for so every set up through the one containing has a square
大提示：
过了 后间距超过 ，所以剩下每组至多含一个平方数；数出不超过 的平方数再相减
Past the gaps exceed so each remaining set holds at most one square; count the squares up to and subtract
解答：
相邻平方数 与 满足差值关系 的条件恰好是 ，所以从 到 的平方数不会跳过任何一个百数段：每个集合 都含有完全平方数。对 ，间距 超过 ，所以集合 中每个至多含一个平方数。
涉及的最大数为 ，且 。因此落在 中的平方数是 ，共 个平方数，占据 个不同集合，而这部分共有 个集合。
因此有 个集合不含完全平方数。
Consecutive squares and differ by for so the squares from to never skip a hundred-block: every set contains a perfect square. For the gap exceeds so each of the sets contains at most one square.
The largest number involved is and So the squares landing in are — that is, squares occupying distinct sets out of those
Therefore sets contain no perfect square.
8.
求正整数 ，使得
Find the positive integer such that
小提示：
每次合并两项，使用 （这里所有乘积 都很小）
Combine two terms at a time using (valid here since all products are small)
大提示：
前三个反正切之和为 ；最后一项必须把它补到
The first three arctangents sum to the last term must top it up to
解答：
对于正数 ，当 时，正切加法公式给出 。应用两次：
方程变为 ，所以 。清除分母得 ，因此 。
For positive with the tangent addition formula gives Applying it twice:
The equation becomes so Clearing denominators, giving
9.
十个相同的木箱尺寸均为 英尺 英尺 英尺。第一个木箱平放在地板上。其余九个木箱依次平放在前一个木箱上方，且每个木箱的朝向随机选择。设 为这堆木箱高度恰好达到 英尺的概率，其中 和 为互质正整数。求 。
Ten identical crates each have dimensions ft ft ft. The first crate is placed flat on the floor. Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random. Let be the probability that the stack of crates is exactly ft tall, where and are relatively prime positive integers. Find
小提示：
若 、、 个木箱分别贡献高度 、、，则 且
If crates contribute heights then and
大提示：
相减得 有三个解；在 个等可能堆法中，用多项式系数数出每个解的排列数
Subtracting gives with three solutions; count the orderings of each with multinomial coefficients out of equally likely stacks
解答：
每个木箱独立地贡献高度 、 或 ，概率各为 ，所以共有 个等可能堆法。若 、、 个木箱的高度分别为 、、，则 且 ；减去第一个方程的三倍，得 ，所以
这些分别可排列为 、 和 种，共 种堆法。概率为 。因为 ，所以这已是最简分数。因此 。
Each crate independently contributes height or each with probability so there are equally likely stacks. If crates have heights then and subtracting three times the first equation gives so
These can be ordered in and ways, for stacks in all. The probability is which is in lowest terms since Thus
10.
设 为等腰梯形，，且较长底边 处的角为 。两条对角线的长度为 。点 到顶点 和 的距离分别为 和 。令 为从 到 的高的垂足。距离 可写成 的形式，其中 和 为正整数，且 不被任何质数的平方整除。求 。
Let be an isosceles trapezoid with whose angle at the longer base is The diagonals have length and point is at distances and from vertices and respectively. Let be the foot of the altitude from to The distance can be expressed in the form where and are positive integers and is not divisible by the square of any prime. Find
小提示：
三角不等式给出 ，所以
The triangle inequality gives so
大提示：
在三角形 中用正弦定理得 ，所以处处取等：，且 在射线 上、位于 的外侧
The Law of Sines in triangle gives so equality holds everywhere: and lies on ray beyond
解答：
由三角不等式， ，所以 。另一方面，在三角形 中， 处的角为 ，且 ，所以正弦定理给出
两个界迫使 ，所以 ，并且三角不等式取等说明 在直线 上，且 在 与 之间。由直角三角形， ，又因为 ，垂足满足 。
点 和 在直线 上且位于 的同侧，所以 ，因此 。
By the triangle inequality, so On the other hand, in triangle the angle at is and so the Law of Sines gives
Both bounds force so and equality in the triangle inequality means lies on line with between and From the right triangle, and since the foot satisfies
Points and are on line on the same side of so and
11.
考虑只由 和 组成的序列，并且它满足：每一段连续的 的长度都是偶数，每一段连续的 的长度都是奇数。这样的序列例子有 、 和 ，而 不是这样的序列。长度为 的这种序列有多少个？
Consider sequences that consist entirely of ’s and ’s and that have the property that every run of consecutive ’s has even length, and every run of consecutive ’s has odd length. Examples of such sequences are and while is not such a sequence. How many such sequences have length
小提示：
按第一个字母分类：以 开头的合法序列必须以 开始，后面接任意合法序列（可以为空）
Classify by the first letter: a valid sequence starting with must begin with followed by any valid (possibly empty) sequence
大提示：
设 分别计数长度为 且以 或 开头的合法序列：，。迭代到 。
With counting valid length- sequences starting with or and Iterate up to
解答：
设 和 分别表示长度为 、以 开头和以 开头的合法序列数。以 开头的序列先是 ，后面接一个长度为 的任意合法序列（可以为空），所以 ，其中空序列计为一个。以 开头的序列要么先是单个 ，后面接以 开头的序列；要么先是 ，后面接以 开头的序列，所以 。
从 和 开始， 在 时依次为
长度为 的合法序列数为 。
Let and count valid sequences of length beginning with and with A sequence beginning with starts with followed by any valid sequence of length (possibly empty), so where the empty sequence counts once. A sequence beginning with starts either with a single followed by a sequence beginning with or with followed by a sequence beginning with so
Starting from and the pairs for are
The number of valid sequences of length is
12.
在一段很长的笔直单向单车道公路上，所有汽车都以相同速度行驶，并且都遵守安全规则：前车车尾到后车车头的距离，按速度每 千米/小时或其不足部分为一车长来计算。（因此，一辆以 千米/小时行驶的汽车，其车头会在前车车尾后方四个车长处。）
路边的光电传感器统计一小时内经过的汽车数量。假设每辆汽车长 米，且汽车可以以任意速度行驶。令 为一小时内能经过该光电传感器的最大整数辆汽车数。求 除以 的商。
On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance from the back of the car ahead to the front of the car behind is exactly one car length for each kilometers per hour of speed or fraction thereof. (Thus the front of a car traveling kilometers per hour will be four car lengths behind the back of the car in front of it.)
A photoelectric eye by the side of the road counts the number of cars that pass in one hour. Assuming that each car is meters long and that the cars can travel at any speed, let be the maximum whole number of cars that can pass the photoelectric eye in one hour. Find the quotient when is divided by
小提示：
速度为 千米/小时时，车头到车头的间距为 米，而每小时有 米长的车流经过传感器
At km/h the front-to-front spacing is meters, and meters of traffic pass the eye per hour
大提示：
对 ，间隔数趋近于一个永远达不到的极限；但若一小时开始时正有一辆车在传感器处，完整经过的汽车数可以达到这个极限
For the gap count rises toward a limit it never attains, but with a car exactly at the eye as the hour starts, the number of whole cars passing does reach that limit
解答：
设汽车速度为 千米/小时。安全间隔为 个车长，所以相邻车头相距 米，一小时内有 米长的车流经过传感器，也就是每小时 个间隔。
固定 时， 在 处最大，此时等于 。它总是小于 ，但会趋近于 ； 越大就越接近。虽然间隔数永远达不到 ，汽车数可以达到：取足够大的 ，使经过的间隔数超过 ，并让计时开始时恰有一辆车在传感器处。这辆车加上随后 个完整间隔各对应的一辆车，共 辆车。
所以 ， 除以 的商为 。
Suppose the cars travel at kilometers per hour. The gap is car lengths, so successive fronts are meters apart, and in one hour a column of meters of traffic passes the eye — that is, gaps per hour.
For a fixed value the count is largest at where it equals This is always less than but approaches as grows. Although the gap count never reaches the car count can: choose so large that more than gaps pass, and start the hour with a car exactly at the eye. That car, plus one car for each of the complete gaps that follow, makes cars.
So and the quotient when is divided by is
13.
令 假设
存在一点 ，使得对所有这样的多项式都有 ，其中 、、 为正整数， 与 互质，且 。求 。
Let Suppose that
There is a point for which for all such polynomials, where and are positive integers, and are relatively prime, and Find
小提示：
依次应用条件：前五个条件迫使 、、
Apply the conditions in order: the first five force
大提示：
最终 ；对任意 和 都成立的点，必须是 与 的公共零点，而它们可以漂亮地分解
Eventually a point that works for every choice of and must be a common zero of and which factor nicely
解答：
由 得 。将 相加和相减，得 且 ；同样由 得 且 。接着 与 化为 和 ，所以 ，。此时 ，而 给出 ，即 。
因此 一个点 若使该式对所有 都为零，就必须使两个括号都为零。第一个括号分解为 ，所以对新点（）需要 。第二个括号为 ；代入 ，将 化为 其根为 和 。
根 重现了已给点 ，所以新点有 ，且 。因此 ，。
From we get Adding and subtracting gives and similarly give and Then and reduce to and so and Now and gives i.e.
Therefore and a point that is a zero for every choice of must kill both brackets. The first bracket factors as so for a new point (with ) we need The second bracket is substituting turns into whose roots are and
The root reproduces the given point so the new point has and Thus and
14.
设 是圆 的一条直径。将 经过 延长到 。点 在 上，使得直线 与 相切。点 是从 到直线 的垂足。已知 ，令 表示线段 的最大可能长度。求 。
Let be a diameter of circle Extend through to Point lies on so that line is tangent to Point is the foot of the perpendicular from to line Suppose and let denote the maximum possible length of segment Find
小提示：
将圆心放在原点；若 ，切线为
Put the center at the origin; if the tangent line is
大提示：
垂足为 ，而 可化简为关于 的二次式，求其最大值
The foot is and simplifies to a quadratic in — maximize it
解答：
将圆心 放在原点，半径为 ，则 ，。若切点为 ，切线为 ；它与 轴交于 ，而该交点位于 的外侧当且仅当 。记 ，从 到该直线的有向距离为 ，所以垂足为 。
于是 。利用 可得 这个关于 的二次式在 处取最大值，该值在 内（此时 ），给出 。
因此 。
Place the center at the origin with radius so and If the point of tangency is the tangent line is it meets the -axis at which lies beyond exactly when Writing the signed distance from to the line is so the foot of the perpendicular is
Then and using This quadratic in is maximized at which is inside (there ), giving
Therefore
15.
一张正方形纸片的边长为 。从每个角按如下方式剪去一个楔形：在每个角处，该楔形的两条剪线都从距离该角 的位置开始，并在对角线上以 的角相交（见下图）。然后沿连接相邻剪线端点的线将纸片向上折起。当一条剪口的两条边相遇时，将它们粘在一起。所得的是一个纸盘，其侧面与底面不成直角。纸盘的高度，也就是底面所在平面与上边缘形成的平面之间的垂直距离，可写成 的形式，其中 和 为正整数，，且 不被任何质数的 次方整除。求 。
A square piece of paper has sides of length From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at distance from the corner, and they meet on the diagonal at an angle of (see the figure below). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upper edges, can be written in the form where and are positive integers, and is not divisible by the th power of any prime. Find
小提示：
将角设为 ，剪线起点 ；剪线在对角线上交于 ，且 、
Set the corner at with cut start the cuts meet at on the diagonal with and
大提示：
折叠会使 绕过 的折线转到 的上方；高度为 ，其中 在 上方的折线上
Folding swings about the fold line through to a point above the height is with on the fold line above
解答：
将角放在原点 ，两条边沿正坐标轴，记 。底边上的剪线起点为 ，两条剪线交于点 ，它位于对角线 上，且两条剪线各自与对角线成 角。在三角形 中，，，所以正弦定理给出 。折线是过 的水平线和竖直线。令 为水平折线上正好位于 上方的点，令 为过 的竖直线与对角线的交点。因为 ，线段 与底边成 角，所以
当底部纸条沿过 的水平线向上折起时，点 始终以 为转动半径、以 为转动圆心，并在过 、垂直于该折线的竖直平面内移动。由对称性，两条粘合的剪口边在对角线上方相遇，因此 最终落在点 ，它就在 的正上方，而 就是纸盘高度。由勾股定理，
所以高度为 ，故 。
Put the corner at the origin with the two sides along the positive axes, and write The cut on the bottom edge starts at and the two cuts meet at on the diagonal each making a angle with the diagonal. In triangle and so the Law of Sines gives The fold lines are the horizontal and vertical lines through Let be the point of the horizontal fold line directly above and the point where the vertical line through meets the diagonal. Since segment makes a angle with the bottom edge, so
When the bottom strip folds up along the horizontal line through point stays at distance from moving in the vertical plane through perpendicular to that fold line. By symmetry the two taped cut edges meet above the diagonal, so lands at a point directly above and is the height of the tray. By the Pythagorean theorem,
So the height is and