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2007 AIME I 真题

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1.

小于 10610^6 的正完全平方数中，有多少个是 2424 的倍数？

How many positive perfect squares less than 10610^6 are multiples of 24?24?

答案：83
知识点：完全平方数质因数分解倍数
难度评级：1790
小提示：

一个完全平方数是 24=23⋅324 = 2^3 \cdot 3 的倍数，当且仅当它的平方根是 1212 的倍数

A perfect square is a multiple of 24=23⋅324 = 2^3 \cdot 3 exactly when its square root is a multiple of 1212

大提示：

统计 1212 的倍数中平方小于 10610^6 的数，也就是 1212 的倍数中小于 10001000 的数

Count the multiples of 1212 whose squares are less than 106,10^6, i.e. the multiples of 1212 less than 10001000

解答：

因为 24=23⋅324 = 2^3 \cdot 3，平方数 N2N^2 是 2424 的倍数，当且仅当 NN 是 1212 的倍数：因子 232^3 要求 NN 含有 222^2，因子 33 要求 NN 含有 33，反过来 (12m)2=144m2(12m)^2 = 144m^2 一定是 2424 的倍数。同时 N2<106N^2 \lt 10^6 当且仅当 N<1000N \lt 1000。

这些 1212 的倍数若小于 10001000，就是 12,24,…,99612, 24, \ldots, 996，共有 99612=83\frac{996}{12} = 83 个。

Since 24=23⋅3,24 = 2^3 \cdot 3, a square N2N^2 is a multiple of 2424 exactly when NN is a multiple of 12:12: the factor 232^3 forces NN to contain 22,2^2, the factor 33 forces NN to contain 3,3, and conversely (12m)2=144m2(12m)^2 = 144m^2 is always a multiple of 24.24. Also N2<106N^2 \lt 10^6 exactly when N<1000.N \lt 1000.

The multiples of 1212 less than 10001000 are 12,24,…,996,12, 24, \ldots, 996, and there are 99612=83\frac{996}{12} = 83 of them.

2.

一条 100100 英尺长的自动人行道以每秒 66 英尺的恒定速度移动。Al 走到人行道起点并站在上面。Bob 在两秒后走到人行道起点，并以每秒 44 英尺的恒定速度沿人行道向前步行。再过两秒，Cy 到达人行道起点，并在人行道旁以每秒 88 英尺的恒定速度快步向前走。在某个时刻，这三人中的一人恰好位于另外两人的正中间。在那个时刻，求中间那人与人行道起点之间的距离，单位为英尺。

A 100100 foot long moving walkway moves at a constant rate of 66 feet per second. Al steps onto the start of the walkway and stands. Bob steps onto the start of the walkway two seconds later and strolls forward along the walkway at a constant rate of 44 feet per second. Two seconds after that, Cy reaches the start of the walkway and walks briskly forward beside the walkway at a constant rate of 88 feet per second. At a certain time, one of these three persons is exactly halfway between the other two. At that time, find the distance in feet between the start of the walkway and the middle person.

答案：52
知识点：路程、速度与时间一次方程分类讨论
难度评级：2020
小提示：

从 Al 踏上人行道开始计时为 tt：Al 在 6t6t，Bob 在 10(t−2)10(t-2)，Cy 在 8(t−4)8(t-4)

Measure time tt from Al’s step: Al is at 6t,6t, Bob at 10(t−2),10(t-2), and Cy at 8(t−4)8(t-4)

大提示：

分别尝试三人中谁在中间：中间位置的两倍必须等于另外两人位置之和。只有一种情形给出 t≥4t \ge 4。

Try each person as the one in the middle: twice the middle position must equal the sum of the other two. Only one case gives t≥4.t \ge 4.

解答：

从 Al 踏上人行道开始，设经过的时间为 tt 秒。Al 站在人行道上，所以位置是 6t6t；Bob 的速度为 6+4=106 + 4 = 10 英尺每秒，所以位置是 10(t−2)10(t - 2)；Cy 在人行道旁以 88 英尺每秒行走，所以位置是 8(t−4)8(t - 4)。当 t≥4t \ge 4 时三人都已经在运动。

中间人的位置的两倍必须等于另外两人的位置之和。若 Bob 在中间，20(t−2)=6t+8(t−4)20(t-2) = 6t + 8(t-4) 得 t=43<4t = \frac{4}{3} \lt 4，不可能。若 Cy 在中间，16(t−4)=6t+10(t−2)16(t-4) = 6t + 10(t-2) 化简为 −64=−20-64 = -20，无解。若 Al 在中间，12t=10(t−2)12t = 10(t-2) +8(t−4)=18t−52+ 8(t-4) = 18t - 52，所以 t=263t = \frac{26}{3}。

此时 Al 的位置为 6⋅263=526 \cdot \frac{26}{3} = 52 英尺，而 Bob 和 Cy 的位置分别为 2003\frac{200}{3} 和 1123\frac{112}{3}，二者的平均值确实是 5252。中间人与起点相距 5252 英尺。

Measure time tt in seconds from when Al steps on. Al stands on the walkway, so he is at 6t;6t; Bob moves at 6+4=106 + 4 = 10 feet per second, so he is at 10(t−2);10(t - 2); Cy walks beside the walkway at 88 feet per second, so he is at 8(t−4).8(t - 4). All three are moving once t≥4.t \ge 4.

The middle person’s position doubled must equal the sum of the other two. If Bob were in the middle, 20(t−2)=6t+8(t−4)20(t-2) = 6t + 8(t-4) gives t=43<4,t = \frac{4}{3} \lt 4, impossible. If Cy were in the middle, 16(t−4)=6t+10(t−2)16(t-4) = 6t + 10(t-2) reduces to −64=−20,-64 = -20, with no solution. If Al is in the middle, 12t=10(t−2)12t = 10(t-2) +8(t−4)=18t−52,+ 8(t-4) = 18t - 52, so t=263.t = \frac{26}{3}.

At that moment Al is at 6⋅263=526 \cdot \frac{26}{3} = 52 feet, while Bob and Cy are at 2003\frac{200}{3} and 1123,\frac{112}{3}, whose average is indeed 52.52. The middle person is 5252 feet from the start.

3.

复数 zz 等于 9+bi9 + bi，其中 bb 是正实数且 i2=−1i^2 = -1。已知 z2z^2 和 z3z^3 的虚部相等，求 bb。

The complex number zz is equal to 9+bi,9 + bi, where bb is a positive real number and i2=−1.i^2 = -1. Given that the imaginary parts of z2z^2 and z3z^3 are equal, find b.b.

答案：15
知识点：复数二项式定理
难度评级：1970
小提示：

用二项式定理展开 (9+bi)2(9 + bi)^2 和 (9+bi)3(9 + bi)^3 然后取出虚部

Expand (9+bi)2(9 + bi)^2 and (9+bi)3(9 + bi)^3 with the binomial theorem and extract the imaginary parts

大提示：

两个虚部都能被 bb 整除；先约去它再求解

Both imaginary parts are divisible by b;b; cancel it before solving

解答：

由二项式定理，z2=(81−b2)+18biz^2 = (81 - b^2) + 18bi，且 z3=(729−27b2)z^3 = (729 - 27b^2) +(243b−b3)i+ (243b - b^3)i。令虚部相等，得到 18b=243b−b318b = 243b - b^3。

因为 bb 为正，所以可同除以 bb，得 b2=243−18=225b^2 = 243 - 18 = 225，因此 b=15b = 15。

By the binomial theorem, z2=(81−b2)+18biz^2 = (81 - b^2) + 18bi and z3=(729−27b2)z^3 = (729 - 27b^2) +(243b−b3)i.+ (243b - b^3)i. Setting the imaginary parts equal gives 18b=243b−b3.18b = 243b - b^3.

Since bb is positive we may divide by b,b, leaving b2=243−18=225,b^2 = 243 - 18 = 225, so b=15.b = 15.

4.

三颗行星绕一颗恒星在同一平面内做圆周运动，恒星位于圆心。所有行星沿同一方向、以各自恒定速度运行，它们的公转周期分别为 6060，8484，和 140140 年。现在恒星和三颗行星的位置共线。它们下一次共线将在 nn 年后。求 nn。

Three planets revolve about a star in coplanar circular orbits with the star at the center. All planets revolve in the same direction, each at a constant speed, and the periods of their orbits are 60,60, 84,84, and 140140 years. The positions of the star and all three planets are currently collinear. They will next be collinear after nn years. Find n.n.

答案：105
知识点：最小公倍数相对速度
难度评级：2230
小提示：

四个天体共线当且仅当每一对行星的公转圈数之差都是 12\frac{1}{2} 的倍数

All four bodies are collinear exactly when each pair of planets has revolution counts differing by a multiple of 12\frac{1}{2}

大提示：

两个两两间隔以每年 1210\frac{1}{210} 圈增长，第三个间隔增长速度是它的两倍；让每个间隔都是 12\frac{1}{2} 的倍数

Two of the pairwise gaps grow at 1210\frac{1}{210} revolution per year and the third twice as fast; make every gap a multiple of 12\frac{1}{2}

解答：

四个天体在同一直线上，当且仅当每一对行星都与恒星共线，也就是说每一对行星的角位置相差 180∘180^\circ 的倍数，即半圈的倍数。在 nn 年内，三颗行星分别转过 n60\frac{n}{60}，n84\frac{n}{84}，和 n140\frac{n}{140} 圈，所以两两差为 n60−n84=n210,n84−n140=n210,n60−n140=n105。\begin{aligned} &\frac{n}{60} - \frac{n}{84} = \frac{n}{210}, \\ &\frac{n}{84} - \frac{n}{140} = \frac{n}{210}, \\ &\frac{n}{60} - \frac{n}{140} = \frac{n}{105} \end{aligned}\text{。}

我们需要 n210\frac{n}{210} 和 n105\frac{n}{105} 都是 12\frac{1}{2} 的倍数。第一个条件要求 nn 是 105105 的倍数，而任何这样的 nn 都会使 n105\frac{n}{105} 为整数。最小的正数选择为 n=105n = 105。

All four bodies lie on one line exactly when every pair of planets is collinear with the star, i.e. when each pair’s angular positions differ by a multiple of 180∘180^\circ — half a revolution. In nn years the planets complete n60,\frac{n}{60}, n84,\frac{n}{84}, and n140\frac{n}{140} revolutions, so the pairwise differences are n60−n84=n210,n84−n140=n210,n60−n140=n105.\begin{aligned} &\frac{n}{60} - \frac{n}{84} = \frac{n}{210}, \\ &\frac{n}{84} - \frac{n}{140} = \frac{n}{210}, \\ &\frac{n}{60} - \frac{n}{140} = \frac{n}{105}. \end{aligned}

We need n210\frac{n}{210} and n105\frac{n}{105} to be multiples of 12.\frac{1}{2}. The first requires nn to be a multiple of 105,105, and any such nn makes n105\frac{n}{105} an integer. The smallest positive choice is n=105.n = 105.

5.

将华氏温度 FF 转换为对应摄氏温度 CC 的公式是 C=59(F−32)C = \frac{5}{9}(F - 32)。一个整数华氏温度先被转换为摄氏温度并四舍五入到最接近的整数；得到的整数摄氏温度再转换回华氏温度，并四舍五入到最接近的整数。对于整数华氏温度 TT，若 32≤T≤100032 \le T \le 1000，有多少个原温度等于最终温度？

The formula for converting a Fahrenheit temperature FF to the corresponding Celsius temperature CC is C=59(F−32).C = \frac{5}{9}(F - 32). An integer Fahrenheit temperature is converted to Celsius and rounded to the nearest integer; the resulting integer Celsius temperature is converted back to Fahrenheit and rounded to the nearest integer. For how many integer Fahrenheit temperatures TT with 32≤T≤100032 \le T \le 1000 does the original temperature equal the final temperature?

答案：539
知识点：取整函数模运算小情形
难度评级：2560
小提示：

将 99 加到 FF 上，恰好会把 55 加到 59(F−32)\frac{5}{9}(F - 32) 上，所以 TT 与 T+9T + 9 的往返转换行为相同

Adding 99 to FF adds exactly 55 to 59(F−32),\frac{5}{9}(F - 32), so the round trip behaves identically for TT and T+9T + 9

大提示：

手算 T=32T = 32 到 4040，看每连续 99 个温度中有多少个能保留下来，再单独处理 995995 到 10001000

Check T=32T = 32 through 4040 by hand to see how many of each 99 consecutive temperatures survive, then handle 995995 through 10001000 separately

解答：

将 99 加到 FF 上，恰好会把 55 加到 59(F−32)\frac{5}{9}(F - 32) 上，因而四舍五入后的摄氏值也增加 55，最终华氏值也增加 99。所以 TT 能回到自身当且仅当 T+9T + 9 能回到自身，因此只需检查连续九个温度。检查 3232 到 4040：最终值为 32,34,34,36,36,37,37,39,3932, 34, 34, 36, 36, 37, 37, 39, 39，所以恰好有 32,34,36,37,3932, 34, 36, 37, 39 这五个温度能保留下来。

从 3232 到 994994 的范围含有 963=107⋅9963 = 107 \cdot 9 个整数，贡献 107⋅5=535107 \cdot 5 = 535 个保留下来的温度。剩下的 995,…,1000995, \ldots, 1000 行为与 32,…,3732, \ldots, 37 相同，其中 32,34,36,3732, 34, 36, 37 能保留下来，又增加 44 个。

总数为 535+4=539535 + 4 = 539。

Adding 99 to FF adds exactly 55 to 59(F−32),\frac{5}{9}(F - 32), hence 55 to the rounded Celsius value, hence 99 to the final Fahrenheit value. So TT returns to itself if and only if T+9T + 9 does, and it suffices to check nine consecutive temperatures. Checking 3232 through 40:40: the final values are 32,34,34,36,36,37,37,39,39,32, 34, 34, 36, 36, 37, 37, 39, 39, so exactly the five temperatures 32,34,36,37,3932, 34, 36, 37, 39 survive.

The range from 3232 through 994994 contains 963=107⋅9963 = 107 \cdot 9 integers, contributing 107⋅5=535107 \cdot 5 = 535 survivors. The remaining 995,…,1000995, \ldots, 1000 behave like 32,…,37,32, \ldots, 37, of which 32,34,36,3732, 34, 36, 37 survive, adding 44 more.

The total is 535+4=539.535 + 4 = 539.

6.

一只青蛙被放在数轴原点，并按如下规则移动：每次移动时，青蛙前进到坐标更大的整数点中最近的 33 的倍数，或前进到坐标更大的整数点中最近的 1313 的倍数。一个移动序列是由合法移动对应的坐标组成的序列，从 00 开始，以 3939 结束。例如，00、33、66、1313、1515、2626、3939 是一个移动序列。青蛙可能有多少个移动序列？

A frog is placed at the origin on the number line, and moves according to the following rule: in a given move, the frog advances to either the closest point with a greater integer coordinate that is a multiple of 3,3, or to the closest point with a greater integer coordinate that is a multiple of 13.13. A move sequence is a sequence of coordinates which correspond to valid moves, beginning with 0,0, and ending with 39.39. For example, 0,0, 3,3, 6,6, 13,13, 15,15, 26,26, 3939 is a move sequence. How many move sequences are possible for the frog?

答案：169
知识点：乘法原理分类讨论
难度评级：2430
小提示：

在标志点 1313 和 2626 处分段：分别计算每一段的路径数并相乘

Split the journey at the landmarks 1313 and 26:26: count routes for each segment and multiply

大提示：

有 55 条路径连接相邻的 1313 的倍数，而有 44 条路径跨过一段并跳过中间那个 1313 的倍数

There are 55 routes between consecutive multiples of 13,13, and 44 routes across a stretch that skips the multiple of 1313 in the middle

解答：

在标志点 1313 和 2626 处分段。从 00 出发时，青蛙沿 33 的倍数上升，并可跳到 1313，起跳点可为 00、33、66、99、1212 中的任一点，所以共有 55 条从 00 到 1313 的路径；同理，有 55 条从 1313 到 2626 的路径（跳到 2626 时可从 1313、1515、1818、2121、2424 起跳），也有 55 条从 2626 到 3939 的路径。若完全跳过 1313，青蛙必须每次选择 33 的倍数选项经过 12→1512 \to 15，然后跳到 2626，起跳点为 1515、1818、2121、2424 中的一点：这给出 44 条从 00 到 2626 且避开 1313 的路径。同理，有 44 条从 1313 到 3939 且避开 2626 的路径，也有 44 条从 00 到 3939 且同时避开二者的路径。

合并各段：经过两个标志点的路径数为 5⋅5⋅5=1255 \cdot 5 \cdot 5 = 125；只经过 1313 的为 5⋅4=205 \cdot 4 = 20；只经过 2626 的为 4⋅5=204 \cdot 5 = 20；两者都不经过的为 44。总数为 125+20+20+4=169125 + 20 + 20 + 4 = 169。

Split the journey at the landmarks 1313 and 26.26. From 00 the frog climbs the multiples of 33 and may jump to 1313 from any of 0,0, 3,3, 6,6, 9,9, 12,12, giving 55 routes from 00 to 13;13; likewise there are 55 routes from 1313 to 2626 (jump to 2626 from 13,13, 15,15, 18,18, 21,21, 2424) and 55 from 2626 to 39.39. To skip 1313 entirely the frog must take the multiple-of-33 option every time through 12→15,12 \to 15, then jump to 2626 from one of 15,15, 18,18, 21,21, 24:24: 44 routes from 00 to 2626 avoiding 13.13. Similarly there are 44 routes from 1313 to 3939 avoiding 26,26, and 44 from 00 to 3939 avoiding both.

Combining the segments: through both landmarks, 5⋅5⋅5=125;5 \cdot 5 \cdot 5 = 125; through 1313 only, 5⋅4=20;5 \cdot 4 = 20; through 2626 only, 4⋅5=20;4 \cdot 5 = 20; through neither, 4.4. The total is 125+20+20+4=169.125 + 20 + 20 + 4 = 169.

7.

令 N=∑k=11000k⋅(⌈log⁡2k⌉−⌊log⁡2k⌋)。\begin{aligned} N &= \sum_{k=1}^{1000} k \\ &\quad {}\cdot \left(\lceil \log_{\sqrt{2}} k \rceil - \lfloor \log_{\sqrt{2}} k \rfloor\right) \end{aligned}\text{。}求 NN 除以 10001000 的余数。（这里 ⌊x⌋\lfloor x \rfloor 表示小于或等于 xx 的最大整数，⌈x⌉\lceil x \rceil 表示大于或等于 xx 的最小整数。）

Let N=∑k=11000k⋅(⌈log⁡2k⌉−⌊log⁡2k⌋).\begin{aligned} N &= \sum_{k=1}^{1000} k \\ &\quad {}\cdot \left(\lceil \log_{\sqrt{2}} k \rceil - \lfloor \log_{\sqrt{2}} k \rfloor\right). \end{aligned} Find the remainder when NN is divided by 1000.1000. (Here ⌊x⌋\lfloor x \rfloor denotes the greatest integer that is less than or equal to x,x, and ⌈x⌉\lceil x \rceil denotes the least integer that is greater than or equal to x.x.)

答案：477
知识点：取整函数对数2的幂求和
难度评级：2410
小提示：

⌈x⌉−⌊x⌋\lceil x \rceil - \lfloor x \rfloor 等于 11，除非 xx 是整数；此时它等于 00

⌈x⌉−⌊x⌋\lceil x \rceil - \lfloor x \rfloor equals 11 unless xx is an integer, when it equals 00

大提示：

对整数 kk，log⁡2k\log_{\sqrt{2}} k 是整数当且仅当 kk 是 22 的幂，所以要从所有整数的和中减去 22 的幂；这里所有整数的和是 1+2+⋯+10001 + 2 + \cdots + 1000

For integers k,k, log⁡2k\log_{\sqrt{2}} k is an integer exactly when kk is a power of 2,2, so subtract the powers of 22 from 1+2+⋯+10001 + 2 + \cdots + 1000

解答：

差 ⌈x⌉−⌊x⌋\lceil x \rceil - \lfloor x \rfloor 在参数不是整数时等于 11，而当 xx 是整数时等于 00。现在，log⁡2k\log_{\sqrt{2}} k 是整数当且仅当 k=(2)jk = (\sqrt{2})^j，其中 jj 为某个整数；要使 kk 为整数，jj 必须为偶数，也就是说 kk 必须是 22 的幂。不超过 10001000 的这些幂为 20,21,…,29=5122^0, 2^1, \ldots, 2^9 = 512。

因此 N=∑k=11000k−∑j=092j=1000⋅10012−1023=500500−1023=499477，\begin{aligned} N &= \sum_{k=1}^{1000} k - \sum_{j=0}^{9} 2^j \\ &= \frac{1000 \cdot 1001}{2} - 1023 \\ &= 500500 - 1023 = 499477 \end{aligned}\text{，}除以 10001000 的余数为 477477。

The difference ⌈x⌉−⌊x⌋\lceil x \rceil - \lfloor x \rfloor equals 11 when xx is not an integer and 00 when it is. Now log⁡2k\log_{\sqrt{2}} k is an integer exactly when k=(2)jk = (\sqrt{2})^j for some integer j,j, and for kk to be an integer, jj must be even — that is, kk must be a power of 2.2. The powers at most 10001000 are 20,21,…,29=512.2^0, 2^1, \ldots, 2^9 = 512.

Therefore N=∑k=11000k−∑j=092j=1000⋅10012−1023=500500−1023=499477,\begin{aligned} N &= \sum_{k=1}^{1000} k - \sum_{j=0}^{9} 2^j \\ &= \frac{1000 \cdot 1001}{2} - 1023 \\ &= 500500 - 1023 = 499477, \end{aligned} and the remainder upon division by 10001000 is 477.477.

8.

多项式 P(x)P(x) 是三次多项式。求 kk 的最大值，使得多项式 Q1(x)=x2+(k−29)x−kQ_1(x) = x^2 + (k - 29)x - k 和 Q2(x)=2x2+(2k−43)x+kQ_2(x) = 2x^2 + (2k - 43)x + k 都是 P(x)P(x) 的因式。

The polynomial P(x)P(x) is cubic. What is the largest value of kk for which the polynomials Q1(x)=x2+(k−29)x−kQ_1(x) = x^2 + (k - 29)x - k and Q2(x)=2x2+(2k−43)x+kQ_2(x) = 2x^2 + (2k - 43)x + k are both factors of P(x)?P(x)?

答案：30
知识点：多项式二次方程因式分解
难度评级：2500
小提示：

一个三次多项式的两个二次因式必须有一个公共根

Two quadratic factors of a cubic must share a common root

大提示：

在公共根 rr 处，2Q1(r)−Q2(r)=02Q_1(r) - Q_2(r) = 0 是一个线性方程，给出 r=−k5r = -\frac{k}{5}；再代回 Q1Q_1

At a common root r,r, 2Q1(r)−Q2(r)=02Q_1(r) - Q_2(r) = 0 is a linear equation giving r=−k5;r = -\frac{k}{5}; substitute back into Q1Q_1

解答：

如果 Q1Q_1 和 Q2Q_2 没有公共根，那么它们的乘积是 44 次多项式，却要整除三次多项式 P(x)P(x)，这是不可能的。所以它们有公共根 rr，且 2Q1(r)−Q2(r)=02Q_1(r) - Q_2(r) = 0。计算得 2Q1(x)−Q2(x)=−15x−3k2Q_1(x) - Q_2(x) = -15x - 3k，因此 r=−k5r = -\frac{k}{5}。

代入 Q1(r)=0Q_1(r) = 0，得到 k225−(k−29)k5−k=0\frac{k^2}{25} - (k - 29)\frac{k}{5} - k = 0；两边乘以 2525 并化简，得 −4k2+120k=0-4k^2 + 120k = 0，所以 k=0k = 0 或 k=30k = 30。

当 k=30k = 30 时，Q1(x)=x2+x−30Q_1(x) = x^2 + x - 30 =(x+6)(x−5)= (x + 6)(x - 5)，且 Q2(x)=2x2+17x+30Q_2(x) = 2x^2 + 17x + 30 =(x+6)(2x+5)= (x + 6)(2x + 5)，二者都整除 P(x)=(x+6)(x−5)(2x+5)P(x) = (x + 6)(x - 5)(2x + 5)。最大值为 3030。

If Q1Q_1 and Q2Q_2 had no common root, their product — of degree 44 — would divide the cubic P(x),P(x), which is impossible. So they share a root r,r, and 2Q1(r)−Q2(r)=0.2Q_1(r) - Q_2(r) = 0. Computing, 2Q1(x)−Q2(x)=−15x−3k,2Q_1(x) - Q_2(x) = -15x - 3k, so r=−k5.r = -\frac{k}{5}.

Substituting into Q1(r)=0Q_1(r) = 0 gives k225−(k−29)k5−k=0;\frac{k^2}{25} - (k - 29)\frac{k}{5} - k = 0; multiplying by 2525 and simplifying yields −4k2+120k=0,-4k^2 + 120k = 0, so k=0k = 0 or k=30.k = 30.

For k=30,k = 30, Q1(x)=x2+x−30Q_1(x) = x^2 + x - 30 =(x+6)(x−5)= (x + 6)(x - 5) and Q2(x)=2x2+17x+30Q_2(x) = 2x^2 + 17x + 30 =(x+6)(2x+5),= (x + 6)(2x + 5), and both divide P(x)=(x+6)(x−5)(2x+5).P(x) = (x + 6)(x - 5)(2x + 5). The largest value is 30.30.

9.

在直角三角形 ABCABC 中，直角在 CC，且 CA=30CA = 30，CB=16CB = 16。将两条直角边 CA‾\overline{CA} 和 CB‾\overline{CB} 分别越过 AA 和 BB 延长。点 O1O_1 和 O2O_2 位于三角形外部，并且是两个半径相等的圆的圆心。以 O1O_1 为圆心的圆与斜边以及直角边 CACA 的延长线相切；以 O2O_2 为圆心的圆与斜边以及直角边 CBCB 的延长线相切；两个圆还彼此外切。任一圆的半径长度可表示为 pq\frac{p}{q}，其中 pp 和 qq 为互素正整数。求 p+qp + q。

In right triangle ABCABC with right angle C,C, CA=30CA = 30 and CB=16.CB = 16. Its legs CA‾\overline{CA} and CB‾\overline{CB} are extended beyond AA and B.B. Points O1O_1 and O2O_2 lie in the exterior of the triangle and are the centers of two circles with equal radii. The circle with center O1O_1 is tangent to the hypotenuse and to the extension of leg CA,CA, the circle with center O2O_2 is tangent to the hypotenuse and to the extension of leg CB,CB, and the circles are externally tangent to each other. The length of the radius of either circle can be expressed as pq,\frac{p}{q}, where pp and qq are relatively prime positive integers. Find p+q.p + q.

答案：737
知识点：相切圆直角三角形三角恒等式
难度评级：2920
小提示：

每个圆都内切于三角形的一个外角，所以从 AA 沿斜边到切点的切线长为 rtan⁡A2r\tan\frac{A}{2}

Each circle is inscribed in an external angle of the triangle, so its tangent length from AA along the hypotenuse is rtan⁡A2r\tan\frac{A}{2}

大提示：

两个切点分割斜边：AB=rtan⁡A2+2r+rtan⁡B2AB = r\tan\frac{A}{2} + 2r + r\tan\frac{B}{2}，其中 tan⁡A2=14\tan\frac{A}{2} = \frac{1}{4}，且 tan⁡B2=35\tan\frac{B}{2} = \frac{3}{5}

The tangency points split the hypotenuse: AB=rtan⁡A2+2r+rtan⁡B2,AB = r\tan\frac{A}{2} + 2r + r\tan\frac{B}{2}, with tan⁡A2=14\tan\frac{A}{2} = \frac{1}{4} and tan⁡B2=35\tan\frac{B}{2} = \frac{3}{5}

解答：

斜边长为 AB=302+162=34AB = \sqrt{30^2 + 16^2} = 34。设 T1T_1 和 T2T_2 分别为两个圆与 ABAB 的切点。两圆半径均为 rr，两个圆心都在直线 ABAB 远离三角形的一侧，所以 O1O2‾\overline{O_1 O_2} 平行于 ABAB，并且 T1T2=O1O2=2rT_1 T_2 = O_1 O_2 = 2r，因为两个圆外切。因此 AB=AT1+2r+T2BAB = AT_1 + 2r + T_2 B。

圆 O1O_1 内切于 AA 处由射线 ABAB 和 CA‾\overline{CA} 越过 AA 的延长线形成的角，该角大小为 180∘−∠A180^\circ - \angle A。因此从 AA 出发的切线长为 AT1=rtan⁡(90∘−A2)AT_1 = \frac{r}{\tan\left(90^\circ - \tfrac{A}{2}\right)} =rtan⁡A2= r \tan\frac{A}{2}。由 sin⁡A=1634\sin A = \frac{16}{34} 和 cos⁡A=3034\cos A = \frac{30}{34}，半角公式给出 tan⁡A2=sin⁡A1+cos⁡A=1664=14\tan\frac{A}{2} = \frac{\sin A}{1 + \cos A} = \frac{16}{64} = \frac{1}{4}，同理 tan⁡B2=3034+16=35\tan\frac{B}{2} = \frac{30}{34 + 16} = \frac{3}{5}。

所以 34=r4+2r+3r5=57r2034 = \frac{r}{4} + 2r + \frac{3r}{5} = \frac{57r}{20}，得 r=68057r = \frac{680}{57}。因为 680=23⋅5⋅17680 = 2^3 \cdot 5 \cdot 17 与 57=3⋅1957 = 3 \cdot 19 没有公因数，所以 p+q=680+57=737p + q = 680 + 57 = 737。

The hypotenuse is AB=302+162=34.AB = \sqrt{30^2 + 16^2} = 34. Let T1T_1 and T2T_2 be the points where the circles touch AB.AB. Both centers lie at distance rr from line ABAB on the side away from the triangle, so O1O2‾\overline{O_1 O_2} is parallel to ABAB and T1T2=O1O2=2r,T_1 T_2 = O_1 O_2 = 2r, since the circles are externally tangent. Thus AB=AT1+2r+T2B.AB = AT_1 + 2r + T_2 B.

Circle O1O_1 is inscribed in the angle at AA between ray ABAB and the extension of CA‾\overline{CA} beyond A,A, which measures 180∘−∠A.180^\circ - \angle A. Its tangent length from AA is therefore AT1=rtan⁡(90∘−A2)AT_1 = \frac{r}{\tan\left(90^\circ - \tfrac{A}{2}\right)} =rtan⁡A2.= r \tan\frac{A}{2}. With sin⁡A=1634\sin A = \frac{16}{34} and cos⁡A=3034,\cos A = \frac{30}{34}, the half-angle formula gives tan⁡A2=sin⁡A1+cos⁡A=1664=14,\tan\frac{A}{2} = \frac{\sin A}{1 + \cos A} = \frac{16}{64} = \frac{1}{4}, and similarly tan⁡B2=3034+16=35.\tan\frac{B}{2} = \frac{30}{34 + 16} = \frac{3}{5}.

So 34=r4+2r+3r5=57r20,34 = \frac{r}{4} + 2r + \frac{3r}{5} = \frac{57r}{20}, giving r=68057.r = \frac{680}{57}. Since 680=23⋅5⋅17680 = 2^3 \cdot 5 \cdot 17 and 57=3⋅1957 = 3 \cdot 19 share no common factor, p+q=680+57=737.p + q = 680 + 57 = 737.

10.

在图示的 6×46 \times 4 方格中，要给 1212 个小方格涂色，而小方格总数为 2424，使每一行有两个被涂色的小方格，每一列有三个被涂色的小方格。设 NN 为满足这一性质的涂色方法数。求 NN 除以 10001000 的余数。

In the 6×46 \times 4 grid shown, 1212 of the 2424 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let NN be the number of shadings with this property. Find the remainder when NN is divided by 1000.1000.

答案：860
知识点：有限制的排列组合分类讨论
难度评级：2990
小提示：

先给第 11 列涂色，再按前两列中都被涂色的行数 kk 分类

Shade column 11 first, then classify by the number kk of rows shaded in both of the first two columns

大提示：

那 kk 个仍为空的行必须在剩下两列中都被涂色；选择第 33 列后，第 44 列就被确定

The kk rows still empty must be shaded in both remaining columns; choosing column 33 then forces column 44

解答：

先在第 11 列中选三行涂色：有 (63)=20\binom{6}{3} = 20 种。令 kk 为第 11 列和第 22 列都被涂色的行数；那么第 22 列可用 (3k)(33−k)\binom{3}{k}\binom{3}{3-k} 种方式选择。前两列涂完后，有 kk 行已经有两个涂色方格，6−2k6 - 2k 行有一个涂色方格，kk 行没有涂色方格。

空行必须在第 33 列和第 44 列中都被涂色。第 33 列取这 kk 个空行，再取 3−k3 - k 个已有一个涂色方格的行，这些行共有 6−2k6 - 2k 个，因此有 (6−2k3−k)\binom{6-2k}{3-k} 种方式；第 44 列随后被确定：它必须覆盖这些空行，并恰好覆盖第 33 列跳过的那些已有一个涂色方格的行。

求和得 N=20∑k=03(3k)⋅(33−k)⋅(6−2k3−k)=20(20+54+18+1)=1860，\begin{aligned} N &= 20\sum_{k=0}^{3} \binom{3}{k} \\ &\quad {}\cdot \binom{3}{3-k} \\ &\quad {}\cdot \binom{6-2k}{3-k} \\ &= 20(20 + 54 + 18 + 1) \\ &= 1860 \end{aligned}\text{，}所以余数为 860860。

Shade three of the six rows in column 1:1: (63)=20\binom{6}{3} = 20 ways. Let kk be the number of rows shaded in both columns 11 and 2;2; column 22 can then be chosen in (3k)(33−k)\binom{3}{k}\binom{3}{3-k} ways. After these two columns, kk rows are complete with two shaded squares, 6−2k6 - 2k rows have one, and kk rows have none.

The empty rows must be shaded in both columns 33 and 4.4. Column 33 takes those kk rows plus 3−k3 - k of the 6−2k6 - 2k singly-shaded rows, in (6−2k3−k)\binom{6-2k}{3-k} ways, and column 44 is then forced: it must cover the empty rows and exactly the singly-shaded rows skipped by column 3.3.

Summing, N=20∑k=03(3k)⋅(33−k)⋅(6−2k3−k)=20(20+54+18+1)=1860,\begin{aligned} N &= 20\sum_{k=0}^{3} \binom{3}{k} \\ &\quad {}\cdot \binom{3}{3-k} \\ &\quad {}\cdot \binom{6-2k}{3-k} \\ &= 20(20 + 54 + 18 + 1) \\ &= 1860, \end{aligned} so the remainder is 860.860.

11.

对每个正整数 pp，令 b(p)b(p) 表示唯一的正整数 kk 使得 ∣k−p∣<12|k - \sqrt{p}| \lt \frac{1}{2}。例如，b(6)=2b(6) = 2，且 b(23)=5b(23) = 5。若 S=∑p=12007b(p)S = \sum_{p=1}^{2007} b(p)，求 SS 除以 10001000 的余数。

For each positive integer p,p, let b(p)b(p) denote the unique positive integer kk such that ∣k−p∣<12.|k - \sqrt{p}| \lt \frac{1}{2}. For example, b(6)=2b(6) = 2 and b(23)=5.b(23) = 5. If S=∑p=12007b(p),S = \sum_{p=1}^{2007} b(p), find the remainder when SS is divided by 1000.1000.

答案：955
知识点：取整函数前n项平方和区间内整数计数
难度评级：2610
小提示：

b(p)=kb(p) = k 当且仅当 k2−k+1≤p≤k2+kk^2 - k + 1 \le p \le k^2 + k，这是含有 2k2k 个 pp 值的一段

b(p)=kb(p) = k exactly when k2−k+1≤p≤k2+k,k^2 - k + 1 \le p \le k^2 + k, a block of 2k2k values of pp

大提示：

对 k≤44k \le 44 的完整区段覆盖到 p≤1980p \le 1980；剩下的每个 pp，直到 20072007，都满足 b(p)=45b(p) = 45

Complete blocks for k≤44k \le 44 cover p≤1980;p \le 1980; every remaining pp up to 20072007 has b(p)=45b(p) = 45

解答：

对正整数 kk，条件 ∣k−p∣<12|k - \sqrt{p}| \lt \frac{1}{2} 意味着 (k−12)2<p<(k+12)2\left(k - \frac{1}{2}\right)^2 \lt p \lt \left(k + \frac{1}{2}\right)^2，对整数 pp 而言，这恰好等价于 k2−k+1≤p≤k2+kk^2 - k + 1 \le p \le k^2 + k。因此，满足 b(p)=kb(p) = k 的取值范围含有 2k2k 个 pp。

因为 442+44=198044^2 + 44 = 1980，区段 k=1,…,44k = 1, \ldots, 44 恰好覆盖所有 p≤1980p \le 1980，贡献 ∑k=144k⋅2k=2⋅44⋅45⋅896=58740。\begin{aligned} \sum_{k=1}^{44} k \cdot 2k &= 2 \cdot \frac{44 \cdot 45 \cdot 89}{6} \\ &= 58740 \end{aligned}\text{。}剩下的 2727 个值 p=1981,…,2007p = 1981, \ldots, 2007 都有 b(p)=45b(p) = 45，又增加 27⋅45=121527 \cdot 45 = 1215。

因此 S=58740+1215=59955S = 58740 + 1215 = 59955，余数为 955955。

For a positive integer k,k, the condition ∣k−p∣<12|k - \sqrt{p}| \lt \frac{1}{2} means (k−12)2<p<(k+12)2,\left(k - \frac{1}{2}\right)^2 \lt p \lt \left(k + \frac{1}{2}\right)^2, which for integers pp is exactly k2−k+1≤p≤k2+k.k^2 - k + 1 \le p \le k^2 + k. So b(p)=kb(p) = k for precisely 2k2k values of p.p.

Since 442+44=1980,44^2 + 44 = 1980, the blocks k=1,…,44k = 1, \ldots, 44 exactly cover p≤1980p \le 1980 and contribute ∑k=144k⋅2k=2⋅44⋅45⋅896=58740.\begin{aligned} \sum_{k=1}^{44} k \cdot 2k &= 2 \cdot \frac{44 \cdot 45 \cdot 89}{6} \\ &= 58740. \end{aligned} The remaining 2727 values p=1981,…,2007p = 1981, \ldots, 2007 each have b(p)=45,b(p) = 45, adding 27⋅45=1215.27 \cdot 45 = 1215.

Thus S=58740+1215=59955,S = 58740 + 1215 = 59955, and the remainder is 955.955.

12.

在等腰三角形 ABCABC 中，AA 位于原点，BB 位于 (20,0)(20, 0)。点 CC 在第一象限，且 AC=BCAC = BC，∠BAC=75∘\angle BAC = 75^\circ。将 △ABC\triangle ABC 绕点 AA 逆时针旋转，直到 CC 的像落在正 yy 轴上。原三角形与旋转后三角形的公共区域面积可写成 p2+q3+r6+sp\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s，其中 pp，qq，rr，ss 为整数。求 p−q+r−s2\frac{p - q + r - s}{2}。

In isosceles triangle ABC,ABC, AA is located at the origin and BB is located at (20,0).(20, 0). Point CC is in the first quadrant with AC=BCAC = BC and ∠BAC=75∘.\angle BAC = 75^\circ. If △ABC\triangle ABC is rotated counterclockwise about point AA until the image of CC lies on the positive yy-axis, the area of the region common to the original triangle and the rotated triangle is in the form p2+q3+r6+s,p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s, where p,p, q,q, r,r, ss are integers. Find p−q+r−s2.\frac{p - q + r - s}{2}.

答案：875
知识点：变换正弦定理相似面积分割
难度评级：3270
小提示：

旋转角为 15∘15^\circ，且由于 ∠ABC=75∘\angle ABC = 75^\circ，AB′AB'，也就是 ABAB 的像，垂直于 BCBC

The rotation is by 15∘,15^\circ, and since ∠ABC=75∘,\angle ABC = 75^\circ, the image AB′AB' of ABAB is perpendicular to BCBC

大提示：

重叠部分为 [AB′F]−[EB′D][AB'F] - [EB'D]，其中 DD 是 AB′AB' 到 BCBC 的垂足，EE、FF 在 B′C′B'C' 上；使用正弦定理以及 △EB′D∼△ABD\triangle EB'D \sim \triangle ABD

The overlap is [AB′F]−[EB′D],[AB'F] - [EB'D], where DD is the foot of AB′AB' on BCBC and E,E, FF lie on B′C′;B'C'; use the law of sines and △EB′D∼△ABD\triangle EB'D \sim \triangle ABD

解答：

因为 ACAC 与正 xx 轴成 75∘75^\circ 角，所以旋转角为 15∘15^\circ。设 B′B' 和 C′C' 分别为 BB 和 CC 的像。由于 ∠B′AB=15∘\angle B'AB = 15^\circ 且 ∠ABC=75∘\angle ABC = 75^\circ，线段 AB′AB' 垂直于 BCBC；令 DD 为它们的交点，并令 E=BC∩B′C′E = BC \cap B'C'，F=AC∩B′C′F = AC \cap B'C'。公共区域为四边形 ADEFADEF，其面积为 [AB′F]−[EB′D][AB'F] - [EB'D]。

在三角形 AB′FAB'F 中，∠FAB′=75∘−15∘=60∘\angle FAB' = 75^\circ - 15^\circ = 60^\circ，且 ∠AB′F=75∘\angle AB'F = 75^\circ，所以 ∠AFB′=45∘\angle AFB' = 45^\circ，正弦定理给出 B′F=20sin⁡60∘sin⁡45∘B'F = \frac{20\sin 60^\circ}{\sin 45^\circ} =106= 10\sqrt{6}。又 sin⁡75∘=6+24\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}，[AB′F]=12⋅20⋅106 sin⁡75∘=50(3+3)。\begin{aligned} [AB'F] &= \tfrac{1}{2} \cdot 20 \cdot 10\sqrt{6}\,\sin 75^\circ \\ &= 50(3 + \sqrt{3}) \end{aligned}\text{。}

在直角三角形 ABDABD 中，AD=20cos⁡15∘AD = 20\cos 15^\circ，且 BD=20sin⁡15∘BD = 20\sin 15^\circ，所以 [ABD]=200sin⁡15∘cos⁡15∘[ABD] = 200\sin 15^\circ\cos 15^\circ =100sin⁡30∘=50= 100\sin 30^\circ = 50，并且 B′D=20(1−cos⁡15∘)B'D = 20(1 - \cos 15^\circ)。三角形 EB′DEB'D 与 ABDABD 相似（都在 DD 处为直角，且 ∠EB′D=∠ABD=75∘\angle EB'D = \angle ABD = 75^\circ），所以利用 cos⁡15∘=6+24\cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}，[EB′D]=50(1−cos⁡15∘sin⁡15∘)2=50⋅(15+83−66−102)。\begin{aligned} [EB'D] &= 50\left(\frac{1 - \cos 15^\circ}{\sin 15^\circ}\right)^2 \\ &= 50 \\ &\quad {}\cdot \left(15 + 8\sqrt{3} - 6\sqrt{6} - 10\sqrt{2}\right) \end{aligned}\text{。}因此，[ADEF]=50(3+3)−50⋅(15+83−66−102)=5002−3503+3006−600，\begin{aligned} [ADEF] &= 50(3 + \sqrt{3}) \\ &\quad {}- 50 \\ &{}\cdot (15 + 8\sqrt{3} - 6\sqrt{6} - 10\sqrt{2}) \\ &= 500\sqrt{2} - 350\sqrt{3} \\ &\quad {}+ 300\sqrt{6} - 600 \end{aligned}\text{，}所以 (p,q,r,s)(p, q, r, s) =(500,−350,300,−600)= (500, -350, 300, -600)，且 p−q+r−s2=17502=875\frac{p - q + r - s}{2} = \frac{1750}{2} = 875。

Since ACAC makes a 75∘75^\circ angle with the positive xx-axis, the rotation is by 15∘.15^\circ. Let B′B' and C′C' be the images of BB and C.C. Because ∠B′AB=15∘\angle B'AB = 15^\circ and ∠ABC=75∘,\angle ABC = 75^\circ, segment AB′AB' is perpendicular to BC;BC; let DD be their intersection, and let E=BC∩B′C′E = BC \cap B'C' and F=AC∩B′C′.F = AC \cap B'C'. The common region is the quadrilateral ADEF,ADEF, whose area is [AB′F]−[EB′D].[AB'F] - [EB'D].

In triangle AB′F,AB'F, ∠FAB′=75∘−15∘=60∘\angle FAB' = 75^\circ - 15^\circ = 60^\circ and ∠AB′F=75∘,\angle AB'F = 75^\circ, so ∠AFB′=45∘,\angle AFB' = 45^\circ, and the law of sines gives B′F=20sin⁡60∘sin⁡45∘B'F = \frac{20\sin 60^\circ}{\sin 45^\circ} =106.= 10\sqrt{6}. With sin⁡75∘=6+24,\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}, [AB′F]=12⋅20⋅106 sin⁡75∘=50(3+3).\begin{aligned} [AB'F] &= \tfrac{1}{2} \cdot 20 \cdot 10\sqrt{6}\,\sin 75^\circ \\ &= 50(3 + \sqrt{3}). \end{aligned}

In right triangle ABD,ABD, AD=20cos⁡15∘AD = 20\cos 15^\circ and BD=20sin⁡15∘,BD = 20\sin 15^\circ, so [ABD]=200sin⁡15∘cos⁡15∘[ABD] = 200\sin 15^\circ\cos 15^\circ =100sin⁡30∘=50,= 100\sin 30^\circ = 50, and B′D=20(1−cos⁡15∘).B'D = 20(1 - \cos 15^\circ). Triangles EB′DEB'D and ABDABD are similar (right angles at D,D, and ∠EB′D=∠ABD=75∘\angle EB'D = \angle ABD = 75^\circ), so, using cos⁡15∘=6+24,\cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}, [EB′D]=50(1−cos⁡15∘sin⁡15∘)2=50⋅(15+83−66−102).\begin{aligned} [EB'D] &= 50\left(\frac{1 - \cos 15^\circ}{\sin 15^\circ}\right)^2 \\ &= 50 \\ &\quad {}\cdot \left(15 + 8\sqrt{3} - 6\sqrt{6} - 10\sqrt{2}\right). \end{aligned} Therefore [ADEF]=50(3+3)−50⋅(15+83−66−102)=5002−3503+3006−600,\begin{aligned} [ADEF] &= 50(3 + \sqrt{3}) \\ &\quad {}- 50 \\ &{}\cdot (15 + 8\sqrt{3} - 6\sqrt{6} - 10\sqrt{2}) \\ &= 500\sqrt{2} - 350\sqrt{3} \\ &\quad {}+ 300\sqrt{6} - 600, \end{aligned} so (p,q,r,s)(p, q, r, s) =(500,−350,300,−600)= (500, -350, 300, -600) and p−q+r−s2=17502=875.\frac{p - q + r - s}{2} = \frac{1750}{2} = 875.

13.

一个以 ABCDABCD 为底面、EE 为顶点的正方形棱锥有八条长为 44 的边。一个平面经过 AE‾\overline{AE}，BC‾\overline{BC}，和 CD‾\overline{CD} 的中点。该平面与棱锥的截面面积可表示为 p\sqrt{p}。求 pp。

A square pyramid with base ABCDABCD and vertex EE has eight edges of length 4.4. A plane passes through the midpoints of AE‾,\overline{AE}, BC‾,\overline{BC}, and CD‾.\overline{CD}. The plane’s intersection with the pyramid has an area that can be expressed as p.\sqrt{p}. Find p.p.

答案：80
知识点：立体几何棱锥坐标几何面积分割
难度评级：3060
小提示：

使用坐标：底面为 (0,0,0)(0,0,0)，(4,0,0)(4,0,0)，(4,4,0)(4,4,0)，(0,4,0)(0,4,0)，顶点为 (2,2,22)(2,2,2\sqrt{2})；经过三个中点的平面为 x+y+22 z=6x + y + 2\sqrt{2}\,z = 6

Use coordinates: base (0,0,0),(0,0,0), (4,0,0),(4,0,0), (4,4,0),(4,4,0), (0,4,0)(0,4,0) and apex (2,2,22);(2,2,2\sqrt{2}); the plane through the three midpoints is x+y+22 z=6x + y + 2\sqrt{2}\,z = 6

大提示：

截面是一个五边形；把它分成一个等腰三角形和一个等腰梯形来求面积

The cross-section is a pentagon; split it into an isosceles triangle and an isosceles trapezoid to find its area

解答：

将底面放在 A=(0,0,0)A = (0,0,0)，B=(4,0,0)B = (4,0,0)，C=(4,4,0)C = (4,4,0)，D=(0,4,0)D = (0,4,0)；则顶点为 E=(2,2,22)E = (2, 2, 2\sqrt{2})，因为 22+22+8=162^2 + 2^2 + 8 = 16。给定的三个中点为 R=(1,1,2)R = (1, 1, \sqrt{2})，S=(4,2,0)S = (4, 2, 0)，和 T=(2,4,0)T = (2, 4, 0)，它们都满足 x+y+22 z=6x + y + 2\sqrt{2}\,z = 6，这是切割平面的方程。

对边 BE‾\overline{BE} 和 DE‾\overline{DE} 参数化，可得平面与它们分别交于 U=(72,12,22)U = \left(\frac{7}{2}, \frac{1}{2}, \frac{\sqrt{2}}{2}\right) 和 V=(12,72,22)V = \left(\frac{1}{2}, \frac{7}{2}, \frac{\sqrt{2}}{2}\right)。截面是五边形 RUSTVRUSTV，其中 RU=RV=7RU = RV = \sqrt{7}，US=VT=3US = VT = \sqrt{3}，ST=22ST = 2\sqrt{2}，对角线 UV=32UV = 3\sqrt{2}。

沿 UV‾\overline{UV} 分割五边形。等腰三角形 RUVRUV 的高为 7−92=52\sqrt{7 - \frac{9}{2}} = \sqrt{\frac{5}{2}}，面积为 12⋅32⋅52=352\frac{1}{2} \cdot 3\sqrt{2} \cdot \sqrt{\frac{5}{2}} = \frac{3\sqrt{5}}{2}。等腰梯形 USTVUSTV 的高为 3−12=52\sqrt{3 - \frac{1}{2}} = \sqrt{\frac{5}{2}}，面积为 12(32+22)52=552\frac{1}{2}(3\sqrt{2} + 2\sqrt{2})\sqrt{\frac{5}{2}} = \frac{5\sqrt{5}}{2}。总面积为 45=804\sqrt{5} = \sqrt{80}，所以 p=80p = 80。

Place the base at A=(0,0,0),A = (0,0,0), B=(4,0,0),B = (4,0,0), C=(4,4,0),C = (4,4,0), D=(0,4,0);D = (0,4,0); the apex is then E=(2,2,22),E = (2, 2, 2\sqrt{2}), since 22+22+8=16.2^2 + 2^2 + 8 = 16. The given midpoints are R=(1,1,2),R = (1, 1, \sqrt{2}), S=(4,2,0),S = (4, 2, 0), and T=(2,4,0),T = (2, 4, 0), and all three satisfy x+y+22 z=6,x + y + 2\sqrt{2}\,z = 6, the equation of the cutting plane.

Parametrizing edges BE‾\overline{BE} and DE‾\overline{DE} shows the plane meets them at U=(72,12,22)U = \left(\frac{7}{2}, \frac{1}{2}, \frac{\sqrt{2}}{2}\right) and V=(12,72,22).V = \left(\frac{1}{2}, \frac{7}{2}, \frac{\sqrt{2}}{2}\right). The cross-section is the pentagon RUSTVRUSTV with RU=RV=7,RU = RV = \sqrt{7}, US=VT=3,US = VT = \sqrt{3}, ST=22,ST = 2\sqrt{2}, and diagonal UV=32.UV = 3\sqrt{2}.

Split the pentagon along UV‾.\overline{UV}. Isosceles triangle RUVRUV has height 7−92=52\sqrt{7 - \frac{9}{2}} = \sqrt{\frac{5}{2}} and area 12⋅32⋅52=352.\frac{1}{2} \cdot 3\sqrt{2} \cdot \sqrt{\frac{5}{2}} = \frac{3\sqrt{5}}{2}. Isosceles trapezoid USTVUSTV has height 3−12=52\sqrt{3 - \frac{1}{2}} = \sqrt{\frac{5}{2}} and area 12(32+22)52=552.\frac{1}{2}(3\sqrt{2} + 2\sqrt{2})\sqrt{\frac{5}{2}} = \frac{5\sqrt{5}}{2}. The total is 45=80,4\sqrt{5} = \sqrt{80}, so p=80.p = 80.

14.

令一个数列定义如下：a1=3a_1 = 3，a2=3a_2 = 3，且对 n≥2n \ge 2，an+1an−1=an2+2007a_{n+1}a_{n-1} = a_n^2 + 2007。求不超过 a20072+a20062a2007a2006\frac{a_{2007}^2 + a_{2006}^2}{a_{2007}a_{2006}} 的最大整数。

Let a sequence be defined as follows: a1=3,a_1 = 3, a2=3,a_2 = 3, and for n≥2,n \ge 2, an+1an−1=an2+2007.a_{n+1}a_{n-1} = a_n^2 + 2007. Find the largest integer less than or equal to a20072+a20062a2007a2006.\frac{a_{2007}^2 + a_{2006}^2}{a_{2007}a_{2006}}.

答案：224
知识点：递推不变量极限情形界定
难度评级：3160
小提示：

将相邻下标的递推式相减，说明 an+1+an−1an\frac{a_{n+1} + a_{n-1}}{a_n} 对所有 nn 都相同

Subtract the relation for consecutive indices to show an+1+an−1an\frac{a_{n+1} + a_{n-1}}{a_n} is the same for all nn

大提示：

将这个常数比例关系乘以 an+1a_{n+1}，并使用原递推式：目标分式等于该常数减去一个很小的正数

Multiply the constant-ratio relation by an+1a_{n+1} and use the original recurrence: the target fraction is that constant minus a tiny positive amount

解答：

对 n≥3n \ge 3 有 an+1an−1=an2+2007a_{n+1}a_{n-1} = a_n^2 + 2007 和 anan−2=an−12+2007a_n a_{n-2} = a_{n-1}^2 + 2007。相减并重新组合，得到 an−1(an+1+an−1)a_{n-1}(a_{n+1} + a_{n-1}) =an(an+an−2)= a_n(a_n + a_{n-2})，因而 an+1+an−1an\frac{a_{n+1} + a_{n-1}}{a_n} 对所有 n≥2n \ge 2 取同一值。因为 a3=32+20073=672a_3 = \frac{3^2 + 2007}{3} = 672，该值为 672+33=225\frac{672 + 3}{3} = 225，数列满足 an+1=225an−an−1a_{n+1} = 225a_n - a_{n-1}。

将 an+1+an−1=225ana_{n+1} + a_{n-1} = 225a_n 乘以 an+1a_{n+1}，并代入 an+1an−1=an2+2007a_{n+1}a_{n-1} = a_n^2 + 2007，得到 an+12+an2+2007=225 anan+1a_{n+1}^2 + a_n^2 + 2007 = 225\,a_n a_{n+1}，因此 an+12+an2an+1an=225−2007anan+1。\frac{a_{n+1}^2 + a_n^2}{a_{n+1}a_n} = 225 - \frac{2007}{a_n a_{n+1}}\text{。}

该数列递增：a3=672>a2a_3 = 672 \gt a_2，且关系 an+1=225an−an−1>ana_{n+1} = 225a_n - a_{n-1} \gt a_n 在每次 an>an−1a_n \gt a_{n-1} 时成立。因此 a2006a2007>6722>2007a_{2006}a_{2007} \gt 672^2 \gt 2007，所以该分式严格介于 224224 与 225225 之间，答案为 224224。

For n≥3,n \ge 3, both an+1an−1=an2+2007a_{n+1}a_{n-1} = a_n^2 + 2007 and anan−2=an−12+2007a_n a_{n-2} = a_{n-1}^2 + 2007 hold. Subtracting and regrouping gives an−1(an+1+an−1)a_{n-1}(a_{n+1} + a_{n-1}) =an(an+an−2),= a_n(a_n + a_{n-2}), so an+1+an−1an\frac{a_{n+1} + a_{n-1}}{a_n} has the same value for every n≥2.n \ge 2. Since a3=32+20073=672,a_3 = \frac{3^2 + 2007}{3} = 672, that value is 672+33=225,\frac{672 + 3}{3} = 225, and the sequence satisfies an+1=225an−an−1.a_{n+1} = 225a_n - a_{n-1}.

Multiplying an+1+an−1=225ana_{n+1} + a_{n-1} = 225a_n by an+1a_{n+1} and substituting an+1an−1=an2+2007a_{n+1}a_{n-1} = a_n^2 + 2007 yields an+12+an2+2007=225 anan+1,a_{n+1}^2 + a_n^2 + 2007 = 225\,a_n a_{n+1}, so an+12+an2an+1an=225−2007anan+1.\frac{a_{n+1}^2 + a_n^2}{a_{n+1}a_n} = 225 - \frac{2007}{a_n a_{n+1}}.

The sequence increases: a3=672>a2,a_3 = 672 \gt a_2, and an+1=225an−an−1>ana_{n+1} = 225a_n - a_{n-1} \gt a_n whenever an>an−1.a_n \gt a_{n-1}. Hence a2006a2007>6722>2007,a_{2006}a_{2007} \gt 672^2 \gt 2007, so the fraction lies strictly between 224224 and 225,225, and the answer is 224.224.

15.

设 ABCABC 为等边三角形，点 DD 和 FF 分别在边 BCBC 和 ABAB 上，且 FA=5FA = 5，CD=2CD = 2。点 EE 在边 CACA 上，使得 ∠DEF=60∘\angle DEF = 60^\circ。三角形 DEFDEF 的面积为 14314\sqrt{3}。边长 ABAB 的两个可能值为 p±qrp \pm q\sqrt{r}，其中 pp 和 qq 为有理数，rr 是不被任何质数平方整除的整数。求 rr。

Let ABCABC be an equilateral triangle, and let DD and FF be points on sides BCBC and AB,AB, respectively, with FA=5FA = 5 and CD=2.CD = 2. Point EE lies on side CACA such that ∠DEF=60∘.\angle DEF = 60^\circ. The area of triangle DEFDEF is 143.14\sqrt{3}. The two possible values of the length of side ABAB are p±qr,p \pm q\sqrt{r}, where pp and qq are rational, and rr is an integer not divisible by the square of a prime. Find r.r.

答案：989
知识点：等边三角形相似三角形面积二次方程
难度评级：3370
小提示：

令 AB=sAB = s 且 AE=tAE = t，从 [ABC][ABC] 中减去三个角上的三角形，可得 5(s−t)+2t=665(s-t) + 2t = 66

Let AB=sAB = s and AE=t,AE = t, and subtract the three corner triangles from [ABC][ABC] to get 5(s−t)+2t=665(s-t) + 2t = 66

大提示：

∠DEF=60∘\angle DEF = 60^\circ 强制 △AEF∼△CDE\triangle AEF \sim \triangle CDE，所以 t(s−t)=10t(s - t) = 10

∠DEF=60∘\angle DEF = 60^\circ forces △AEF∼△CDE,\triangle AEF \sim \triangle CDE, so t(s−t)=10t(s - t) = 10

解答：

令 s=ABs = AB 且 t=AEt = AE。利用 60∘60^\circ 角（分别位于 AA、BB、CC）以及面积公式 12xysin⁡60∘\frac{1}{2}xy\sin 60^\circ：有 [AEF]=34⋅5t[AEF] = \frac{\sqrt{3}}{4} \cdot 5t，[BFD]=34(s−5)(s−2)[BFD] = \frac{\sqrt{3}}{4}(s-5)(s-2)，且 [CDE]=34⋅2(s−t)[CDE] = \frac{\sqrt{3}}{4} \cdot 2(s-t)。从 [ABC]=34s2[ABC] = \frac{\sqrt{3}}{4}s^2 中减去这三个面积并化简，[DEF]=34(5(s−t)+2t−10)=143，\begin{aligned} [DEF] &= \frac{\sqrt{3}}{4}\bigl(5(s - t) + 2t - 10\bigr) \\ &= 14\sqrt{3} \end{aligned}\text{，}所以 5(s−t)+2t=665(s - t) + 2t = 66。

在 EE 处，角 ∠AEF\angle AEF 和 ∠CED\angle CED 之和为 180∘−60∘=120∘180^\circ - 60^\circ = 120^\circ，而在三角形 AEFAEF 中，角 ∠AEF\angle AEF 和 ∠AFE\angle AFE 之和也为 120∘120^\circ。因此 ∠AFE=∠CED\angle AFE = \angle CED，又因为 ∠A=∠C=60∘\angle A = \angle C = 60^\circ，三角形 AEFAEF 与 CDECDE 相似。于是 AEAF=CDCE\frac{AE}{AF} = \frac{CD}{CE} 给出 t5=2s−t\frac{t}{5} = \frac{2}{s - t}，所以 t(s−t)=10t(s - t) = 10。

将 s−t=10ts - t = \frac{10}{t} 代入 5(s−t)+2t=665(s - t) + 2t = 66，得到 50t+2t=66\frac{50}{t} + 2t = 66，即 t2−33t+25=0t^2 - 33t + 25 = 0，所以 t=33±9892t = \frac{33 \pm \sqrt{989}}{2}。由 25t=33−t\frac{25}{t} = 33 - t 可得 10t=25(33−t)\frac{10}{t} = \frac{2}{5}(33 - t)，因此 s=t+10t=3t+665=231±398910s = t + \frac{10}{t} = \frac{3t + 66}{5} = \frac{231 \pm 3\sqrt{989}}{10}。两个值都给出有效构型，所以 r=989r = 989。

Let s=ABs = AB and t=AE.t = AE. Using the 60∘60^\circ angles at A,A, B,B, CC and the area formula 12xysin⁡60∘:\frac{1}{2}xy\sin 60^\circ: [AEF]=34⋅5t,[AEF] = \frac{\sqrt{3}}{4} \cdot 5t, [BFD]=34(s−5)(s−2),[BFD] = \frac{\sqrt{3}}{4}(s-5)(s-2), and [CDE]=34⋅2(s−t).[CDE] = \frac{\sqrt{3}}{4} \cdot 2(s-t). Subtracting all three from [ABC]=34s2[ABC] = \frac{\sqrt{3}}{4}s^2 and simplifying, [DEF]=34(5(s−t)+2t−10)=143,\begin{aligned} [DEF] &= \frac{\sqrt{3}}{4}\bigl(5(s - t) + 2t - 10\bigr) \\ &= 14\sqrt{3}, \end{aligned} so 5(s−t)+2t=66.5(s - t) + 2t = 66.

At E,E, the angles ∠AEF\angle AEF and ∠CED\angle CED sum to 180∘−60∘=120∘,180^\circ - 60^\circ = 120^\circ, while in triangle AEFAEF the angles ∠AEF\angle AEF and ∠AFE\angle AFE also sum to 120∘.120^\circ. Hence ∠AFE=∠CED,\angle AFE = \angle CED, and since ∠A=∠C=60∘,\angle A = \angle C = 60^\circ, triangles AEFAEF and CDECDE are similar. Then AEAF=CDCE\frac{AE}{AF} = \frac{CD}{CE} gives t5=2s−t,\frac{t}{5} = \frac{2}{s - t}, so t(s−t)=10.t(s - t) = 10.

Substituting s−t=10ts - t = \frac{10}{t} into 5(s−t)+2t=665(s - t) + 2t = 66 gives 50t+2t=66,\frac{50}{t} + 2t = 66, or t2−33t+25=0,t^2 - 33t + 25 = 0, so t=33±9892.t = \frac{33 \pm \sqrt{989}}{2}. From 25t=33−t\frac{25}{t} = 33 - t we get 10t=25(33−t),\frac{10}{t} = \frac{2}{5}(33 - t), so s=t+10t=3t+665=231±398910.s = t + \frac{10}{t} = \frac{3t + 66}{5} = \frac{231 \pm 3\sqrt{989}}{10}. Both values yield valid configurations, so r=989.r = 989.