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2001 AIME II 真题

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1.

设 NN 是满足以下性质的最大正整数：从左到右看，NN 的每一对相邻数字都组成一个完全平方数。求 NN 最左边的三位数字。

Let NN be the largest positive integer with the following property: reading from left to right, each pair of consecutive digits of NN forms a perfect square. What are the leftmost three digits of N?N?

答案：816
知识点：数字完全平方数系统列举
难度评级：1890
小提示：

每一对相邻数字必须是两位完全平方数 1616、2525、3636、4949、6464、8181 之一。

Each adjacent digit pair must be one of the two-digit perfect squares 16,16, 25,25, 36,36, 49,49, 64,64, 8181

大提示：

每个数字至多只有一个可能的下一个数字，所以追踪各条链并取最长的一条。

Every digit has at most one possible next digit, so trace the chains and take the longest one

解答：

每一对相邻数字必须是两位完全平方数 1616、2525、3636、4949、6464、8181 之一。因此，若某个数字有后继，它就唯一确定这个后继：1→61 \to 6、2→52 \to 5、3→63 \to 6、4→94 \to 9、6→46 \to 4、8→18 \to 1，而 55 和 99 会结束这个数。

从每个可能的起始数字追踪，最长的数字串为 2525、36493649 和 8164981649。五位链 8→1→6→4→98 \to 1 \to 6 \to 4 \to 9 长于其他所有链，所以 N=81649N = 81649，其最左边三位数字为 816816。

Each pair of consecutive digits must be one of the two-digit squares 16,16, 25,25, 36,36, 49,49, 64,64, 81.81. So each digit determines its successor uniquely if one exists: 1→6,1 \to 6, 2→5,2 \to 5, 3→6,3 \to 6, 4→9,4 \to 9, 6→4,6 \to 4, 8→1,8 \to 1, while 55 and 99 end the number.

Following these chains from each possible starting digit, the longest strings are 25,25, 3649,3649, and 81649.81649. The five-digit chain 8→1→6→4→98 \to 1 \to 6 \to 4 \to 9 beats everything else, so N=81649,N = 81649, whose leftmost three digits are 816.816.

2.

某高中的 20012001 名学生每人都学习西班牙语或法语，也有一些人两种语言都学。学习西班牙语的人数占全校总人数的百分之 8080 到百分之 8585 之间，学习法语的人数占百分之 3030 到百分之 4040 之间。设 mm 为可能同时学习两种语言的最少学生数，MM 为可能同时学习两种语言的最多学生数。求 M−mM - m。

Each of the 20012001 students at a high school studies either Spanish or French, and some study both. The number who study Spanish is between 8080 percent and 8585 percent of the school population, and the number who study French is between 3030 percent and 4040 percent. Let mm be the smallest number of students who could study both languages, and let MM be the largest number of students who could study both languages. Find M−m.M - m.

答案：298
知识点：容斥原理百分数极限情形界定
难度评级：2110
小提示：

若 ss 人学西班牙语、ff 人学法语，则两种都学的人数是 s+f−2001s + f - 2001。

If ss study Spanish and ff study French, then s+f−2001s + f - 2001 students study both

大提示：

百分比范围给出 1601≤s≤17001601 \le s \le 1700 和 601≤f≤800601 \le f \le 800；将 s+fs + f 推到两个极端。

The percentage bounds give 1601≤s≤17001601 \le s \le 1700 and 601≤f≤800;601 \le f \le 800; push s+fs + f to its two extremes

解答：

设学习西班牙语和法语的人数分别为 ss 和 ff。因为每个学生至少学习一种语言，同时学习两种语言的人数为 s+f−2001s + f - 2001。由范围 1600.8<s<1700.851600.8 \lt s \lt 1700.85 得 1601≤s≤17001601 \le s \le 1700，由 600.3<f<800.4600.3 \lt f \lt 800.4 得 601≤f≤800601 \le f \le 800。

重叠人数在 s+fs + f 最小时最小，所以 m=1601+601−2001=201m = 1601 + 601 - 2001 = 201；在 s+fs + f 最大时最大，所以 M=1700+800−2001=499M = 1700 + 800 - 2001 = 499。这两个极端都可以达到，因此 M−m=499−201=298M - m = 499 - 201 = 298。

Let ss and ff be the numbers of students studying Spanish and French. Since every student studies at least one language, the number studying both is s+f−2001.s + f - 2001. The bounds 1600.8<s<1700.851600.8 \lt s \lt 1700.85 force 1601≤s≤1700,1601 \le s \le 1700, and 600.3<f<800.4600.3 \lt f \lt 800.4 force 601≤f≤800.601 \le f \le 800.

The overlap is smallest when s+fs + f is smallest, giving m=1601+601−2001=201,m = 1601 + 601 - 2001 = 201, and largest when s+fs + f is largest, giving M=1700+800−2001=499.M = 1700 + 800 - 2001 = 499. Both extremes are achievable, so M−m=499−201=298.M - m = 499 - 201 = 298.

3.

前四项为 x1=211x_1 = 211、x2=375x_2 = 375、x3=420x_3 = 420 和 x4=523x_4 = 523。对 n≥5n \ge 5，有 xn=xn−1−xn−2+xn−3−xn−4。 \begin{aligned} x_n &= x_{n-1} - x_{n-2} \\ &\quad {}+ x_{n-3} - x_{n-4} \end{aligned}\text{。}求 x531+x753+x975x_{531} + x_{753} + x_{975} 的值。

The first four terms are x1=211,x_1 = 211, x2=375,x_2 = 375, x3=420,x_3 = 420, and x4=523.x_4 = 523. For n≥5,n \ge 5, xn=xn−1−xn−2+xn−3−xn−4. \begin{aligned} x_n &= x_{n-1} - x_{n-2} \\ &\quad {}+ x_{n-3} - x_{n-4}. \end{aligned} Find the value of x531+x753+x975.x_{531} + x_{753} + x_{975}.

答案：898
知识点：递推找规律
难度评级：2070
小提示：

将 xn−1x_{n-1} 的递推式代入 xnx_n 的公式，可得 xn=−xn−5x_n = -x_{n-5}。

Substitute the recurrence for xn−1x_{n-1} into the formula for xnx_n to get xn=−xn−5x_n = -x_{n-5}

大提示：

该数列以 1010 为周期，所以把每个下标模 1010 化简，并用已知值算出 x5x_5。

The sequence repeats with period 10,10, so reduce each subscript mod 1010 and compute x5x_5 from the given values

解答：

当 n≥6n \ge 6 时，将 xn−1x_{n-1} 的递推式代入：xn=(xn−2−xn−3+xn−4−xn−5)−xn−2+xn−3−xn−4=−xn−5。 \begin{aligned} x_n &= \\ &(x_{n-2} - x_{n-3} + x_{n-4} - x_{n-5}) \\ &\quad {}- x_{n-2} \\ &\quad {}+ x_{n-3} - x_{n-4} \\ &= -x_{n-5} \end{aligned}\text{。}因此 xn+10=−xn+5=xnx_{n+10} = -x_{n+5} = x_n，所以数列周期为 1010。

531531、753753 和 975975 除以 1010 的余数分别为 11、33 和 55，所以 x531=x1=211x_{531} = x_1 = 211、x753=x3=420x_{753} = x_3 = 420，且 x975=x5x_{975} = x_5 =x4−x3+x2−x1= x_4 - x_3 + x_2 - x_1 =523−420+375−211= 523 - 420 + 375 - 211 =267= 267。

所求和为 211+420+267=898211 + 420 + 267 = 898。

For n≥6,n \ge 6, substitute the recurrence for xn−1:x_{n-1}: xn=(xn−2−xn−3+xn−4−xn−5)−xn−2+xn−3−xn−4=−xn−5. \begin{aligned} x_n &= \\ &(x_{n-2} - x_{n-3} + x_{n-4} - x_{n-5}) \\ &\quad {}- x_{n-2} \\ &\quad {}+ x_{n-3} - x_{n-4} \\ &= -x_{n-5}. \end{aligned} Hence xn+10=−xn+5=xn,x_{n+10} = -x_{n+5} = x_n, so the sequence has period 10.10.

Since 531,531, 753,753, and 975975 leave remainders 1,1, 3,3, and 55 upon division by 10,10, we get x531=x1=211,x_{531} = x_1 = 211, x753=x3=420,x_{753} = x_3 = 420, and x975=x5x_{975} = x_5 =x4−x3+x2−x1= x_4 - x_3 + x_2 - x_1 =523−420+375−211= 523 - 420 + 375 - 211 =267.= 267.

The sum is 211+420+267=898.211 + 420 + 267 = 898.

4.

设 R=(8,6)R = (8, 6)。方程为 8y=15x8y = 15x 和 10y=3x10y = 3x 的直线分别含有点 PP 和 QQ，且 RR 是 PQ‾\overline{PQ} 的中点。长度 PQPQ 等于 mn\frac{m}{n}，其中 mm 和 nn 是互质正整数。求 m+nm + n。

Let R=(8,6).R = (8, 6). The lines whose equations are 8y=15x8y = 15x and 10y=3x10y = 3x contain points PP and Q,Q, respectively, such that RR is the midpoint of PQ‾.\overline{PQ}. The length PQPQ equals mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

答案：67
知识点：坐标几何中点方程组距离公式
难度评级：2170
小提示：

写成 P=(8t,15t)P = (8t, 15t)、Q=(10u,3u)Q = (10u, 3u)，这样两个直线条件会自动满足。

Write P=(8t,15t)P = (8t, 15t) and Q=(10u,3u)Q = (10u, 3u) so that both line conditions hold automatically

大提示：

中点条件给出 8t+10u=168t + 10u = 16 和 15t+3u=1215t + 3u = 12；解出 tt 和 uu，再用距离公式。

The midpoint condition gives 8t+10u=168t + 10u = 16 and 15t+3u=12;15t + 3u = 12; solve for tt and u,u, then use the distance formula

解答：

两条直线上的点可写成 P=(8t,15t)P = (8t, 15t) 和 Q=(10u,3u)Q = (10u, 3u)。因为 R=(8,6)R = (8, 6) 是 PQ‾\overline{PQ} 的中点，8t+10u=168t + 10u = 16 且 15t+3u=12。15t + 3u = 12\text{。}

第二个方程给出 u=4−5tu = 4 - 5t；代入第一个方程，得 8t+40−50t=168t + 40 - 50t = 16，所以 t=47t = \frac{4}{7}，且 u=87u = \frac{8}{7}。于是 P=(327,607)P = \left(\frac{32}{7}, \frac{60}{7}\right)，Q=(807,247)Q = \left(\frac{80}{7}, \frac{24}{7}\right)。

因而 PQ=(487)2+(367)2PQ = \sqrt{\left(\frac{48}{7}\right)^2 + \left(\frac{36}{7}\right)^2} =12742+32= \frac{12}{7}\sqrt{4^2 + 3^2} =607= \frac{60}{7}，所以 m+n=60+7=67m + n = 60 + 7 = 67。

Points on the two lines can be written P=(8t,15t)P = (8t, 15t) and Q=(10u,3u).Q = (10u, 3u). Since R=(8,6)R = (8, 6) is the midpoint of PQ‾,\overline{PQ}, 8t+10u=168t + 10u = 16 and 15t+3u=12.15t + 3u = 12.

The second equation gives u=4−5t;u = 4 - 5t; substituting into the first, 8t+40−50t=16,8t + 40 - 50t = 16, so t=47t = \frac{4}{7} and u=87.u = \frac{8}{7}. Thus P=(327,607)P = \left(\frac{32}{7}, \frac{60}{7}\right) and Q=(807,247).Q = \left(\frac{80}{7}, \frac{24}{7}\right).

Then PQ=(487)2+(367)2PQ = \sqrt{\left(\frac{48}{7}\right)^2 + \left(\frac{36}{7}\right)^2} =12742+32= \frac{12}{7}\sqrt{4^2 + 3^2} =607,= \frac{60}{7}, so m+n=60+7=67.m + n = 60 + 7 = 67.

5.

一组正数若有三个不同元素可以作为某个面积为正的三角形的边长，则称它具有三角形性质。考虑由连续正整数组成的集合 {4,5,6,…,n}\{4, 5, 6, \ldots, n\}，并且它的每个十元素子集都具有三角形性质。求 nn 的最大可能值。

A set of positive numbers has the triangle property if it has three distinct elements that are the lengths of the sides of a triangle whose area is positive. Consider sets {4,5,6,…,n}\{4, 5, 6, \ldots, n\} of consecutive positive integers, all of whose ten-element subsets have the triangle property. What is the largest possible value of n?n?

答案：253
知识点：三角不等式斐波那契数列极端原理
难度评级：2390
小提示：

在没有三角形的集合中，每个元素至少必须是前两个元素之和。

In a set with no triangle, each element must be at least the sum of the two preceding elements

大提示：

从 4,54, 5 开始的最小十元素无三角形集合按斐波那契式增长：4,5,9,14,23,…4, 5, 9, 14, 23, \ldots。它的最大元素决定 nn 应停在哪里。

The smallest triangle-free ten-element set starting from 4,54, 5 grows like Fibonacci: 4,5,9,14,23,…4, 5, 9, 14, 23, \ldots Its largest element tells you where nn must stop.

解答：

设某个十元素集合 {a1<a2<⋯<a10}\{a_1 \lt a_2 \lt \cdots \lt a_{10}\} 没有三角形。那么任意三个元素都不满足严格三角不等式；特别地，对每个 ii，都有 ai+2≥ai+1+aia_{i+2} \ge a_{i+1} + a_i。从 a1≥4a_1 \ge 4 和 a2≥5a_2 \ge 5 出发，依次迫使 a3≥9a_3 \ge 9、a4≥14a_4 \ge 14、a5≥23a_5 \ge 23、a6≥37a_6 \ge 37、a7≥60a_7 \ge 60、a8≥97a_8 \ge 97、a9≥157a_9 \ge 157，且 a10≥254a_{10} \ge 254。

因此若 n≤253n \le 253，{4,5,…,n}\{4, 5, \ldots, n\} 的任何十元素子集都无法避开三角形，因为它的最大元素至少要达到 254254。反过来，取每一步都等号成立，{4,5,9,14,23,\{4, 5, 9, 14, 23, 37,60,97,157,254}37, 60, 97, 157, 254\} 是 {4,5,…,254}\{4, 5, \ldots, 254\} 中一个没有三角形的子集。

所以最大可能值为 n=253n = 253。

Suppose a ten-element set {a1<a2<⋯<a10}\{a_1 \lt a_2 \lt \cdots \lt a_{10}\} has no triangle. Then every three elements fail the strict triangle inequality; in particular ai+2≥ai+1+aia_{i+2} \ge a_{i+1} + a_i for each i.i. Starting from a1≥4a_1 \ge 4 and a2≥5,a_2 \ge 5, this forces a3≥9,a_3 \ge 9, a4≥14,a_4 \ge 14, a5≥23,a_5 \ge 23, a6≥37,a_6 \ge 37, a7≥60,a_7 \ge 60, a8≥97,a_8 \ge 97, a9≥157,a_9 \ge 157, and a10≥254.a_{10} \ge 254.

So if n≤253,n \le 253, no ten-element subset of {4,5,…,n}\{4, 5, \ldots, n\} can avoid triangles, since its largest element would have to be at least 254.254. Conversely, taking equality throughout, the subset {4,5,9,14,23,\{4, 5, 9, 14, 23, 37,60,97,157,254}37, 60, 97, 157, 254\} of {4,5,…,254}\{4, 5, \ldots, 254\} has no triangle.

Therefore the largest possible value is n=253.n = 253.

6.

正方形 ABCDABCD 内接于一个圆。正方形 EFGHEFGH 的顶点 EE 和 FF 在 CD‾\overline{CD} 上，顶点 GG 和 HH 在圆上。正方形 EFGHEFGH 的面积与正方形 ABCDABCD 的面积之比可表示为 mn\frac{m}{n}，其中 mm 和 nn 是互质正整数且 m<nm \lt n。求 10n+m10n + m。

Square ABCDABCD is inscribed in a circle. Square EFGHEFGH has vertices EE and FF on CD‾\overline{CD} and vertices GG and HH on the circle. The ratio of the area of square EFGHEFGH to the area of square ABCDABCD can be expressed as mn,\frac{m}{n}, where mm and nn are relatively prime positive integers and m<n.m \lt n. Find 10n+m.10n + m.

答案：251
知识点：正方形（几何）圆坐标几何二次方程
难度评级：2390
小提示：

将圆心放在原点；若 ABCDABCD 的边长为 ss，则圆为 x2+y2=s22x^2 + y^2 = \frac{s^2}{2}。

Put the circle’s center at the origin; if ABCDABCD has side s,s, the circle is x2+y2=s22x^2 + y^2 = \frac{s^2}{2}

大提示：

若 EFGHEFGH 的边长为 tt，则 G=(t2,s2+t)G = \left(\frac{t}{2}, \frac{s}{2} + t\right) 在圆上，得到 5t2+4st−s2=05t^2 + 4st - s^2 = 0。

If EFGHEFGH has side t,t, then G=(t2,s2+t)G = \left(\frac{t}{2}, \frac{s}{2} + t\right) lies on the circle, giving 5t2+4st−s2=05t^2 + 4st - s^2 = 0

解答：

将圆心放在原点，设 ABCDABCD 的边长为 ss，于是圆为 x2+y2=s22x^2 + y^2 = \frac{s^2}{2}，边 CD‾\overline{CD} 位于直线 y=s2y = \frac{s}{2} 上。小正方形立在 CD‾\overline{CD} 上，并在 ABCDABCD 外侧：若它的边长为 tt，由对称性 G=(t2,s2+t)G = \left(\frac{t}{2}, \frac{s}{2} + t\right)，且该点必须在圆上。

代入得 t24+(s2+t)2=s22\frac{t^2}{4} + \left(\frac{s}{2} + t\right)^2 = \frac{s^2}{2}，展开为 5t2+4st−s2=05t^2 + 4st - s^2 = 0，即 (5t−s)(t+s)=0(5t - s)(t + s) = 0。因为 t>0t \gt 0，得 t=s5t = \frac{s}{5}。

面积比为 t2s2=125\frac{t^2}{s^2} = \frac{1}{25}，所以 m=1m = 1，n=25n = 25，且 10n+m=25110n + m = 251。

Center the circle at the origin and let ABCDABCD have side s,s, so the circle is x2+y2=s22x^2 + y^2 = \frac{s^2}{2} and side CD‾\overline{CD} lies on the line y=s2.y = \frac{s}{2}. The small square sits on CD‾,\overline{CD}, outside ABCD:ABCD: if its side is t,t, then by symmetry G=(t2,s2+t),G = \left(\frac{t}{2}, \frac{s}{2} + t\right), which must lie on the circle.

Substituting, t24+(s2+t)2=s22,\frac{t^2}{4} + \left(\frac{s}{2} + t\right)^2 = \frac{s^2}{2}, which expands to 5t2+4st−s2=0,5t^2 + 4st - s^2 = 0, or (5t−s)(t+s)=0.(5t - s)(t + s) = 0. Since t>0,t \gt 0, we get t=s5.t = \frac{s}{5}.

The ratio of areas is t2s2=125,\frac{t^2}{s^2} = \frac{1}{25}, so m=1,m = 1, n=25,n = 25, and 10n+m=251.10n + m = 251.

7.

设 △PQR\triangle PQR 是直角三角形，且 PQ=90PQ = 90，PR=120PR = 120，QR=150QR = 150。令 C1C_1 为其内切圆。作 ST‾\overline{ST}，其中 SS 在 PR‾\overline{PR} 上、TT 在 QR‾\overline{QR} 上，使得 ST‾\overline{ST} 垂直于 PR‾\overline{PR} 并与 C1C_1 相切。作 UV‾\overline{UV}，其中 UU 在 PQ‾\overline{PQ} 上、VV 在 QR‾\overline{QR} 上，使得 UV‾\overline{UV} 垂直于 PQ‾\overline{PQ} 并与 C1C_1 相切。令 C2C_2 为 △RST\triangle RST 的内切圆，C3C_3 为 △QUV\triangle QUV 的内切圆。C2C_2 和 C3C_3 的圆心距离可写为 10n\sqrt{10n}。求 nn。

Let △PQR\triangle PQR be a right triangle with PQ=90,PQ = 90, PR=120,PR = 120, and QR=150.QR = 150. Let C1C_1 be the inscribed circle. Construct ST‾,\overline{ST}, with SS on PR‾\overline{PR} and TT on QR‾,\overline{QR}, such that ST‾\overline{ST} is perpendicular to PR‾\overline{PR} and tangent to C1.C_1. Construct UV‾\overline{UV} with UU on PQ‾\overline{PQ} and VV on QR‾\overline{QR} such that UV‾\overline{UV} is perpendicular to PQ‾\overline{PQ} and tangent to C1.C_1. Let C2C_2 be the inscribed circle of △RST\triangle RST and C3C_3 the inscribed circle of △QUV.\triangle QUV. The distance between the centers of C2C_2 and C3C_3 can be written as 10n.\sqrt{10n}. What is n?n?

答案：725
知识点：内切圆、内心与内切圆半径直角三角形相似距离公式
难度评级：2560
小提示：

将 PP 放在原点。大内切圆半径为 90+120−1502=30\frac{90 + 120 - 150}{2} = 30，所以 ST‾\overline{ST} 和 UV‾\overline{UV} 是直线 x=60x = 60 与 y=60y = 60。

Put PP at the origin. The big incircle has radius 90+120−1502=30,\frac{90 + 120 - 150}{2} = 30, so ST‾\overline{ST} and UV‾\overline{UV} are the lines x=60x = 60 and y=60.y = 60.

大提示：

三角形 RSTRST 和 QUVQUV 分别与原三角形相似，比例为 12\frac{1}{2} 和 13\frac{1}{3}；用坐标确定两个小内切圆圆心。

Triangles RSTRST and QUVQUV are similar to the original with ratios 12\frac{1}{2} and 13;\frac{1}{3}; locate each small incircle’s center in coordinates

解答：

直角在 PP，所以令 P=(0,0)P = (0, 0)、Q=(0,90)Q = (0, 90)、R=(120,0)R = (120, 0)。直角三角形的内切圆半径等于两条直角边之和减去斜边后的一半：r1=90+120−1502=30r_1 = \frac{90 + 120 - 150}{2} = 30，所以 C1C_1 的圆心为 (30,30)(30, 30)。在 C1C_1 靠近 RR 的一侧，垂直于 PR‾\overline{PR} 的切线为 x=60x = 60；在靠近 QQ 的一侧，垂直于 PQ‾\overline{PQ} 的切线为 y=60y = 60。

三角形 RSTRST 与三角形 RPQRPQ 相似，比例为 RSRP=60120=12\frac{RS}{RP} = \frac{60}{120} = \frac{1}{2}，所以其内切圆半径为 1515，内切圆 C2C_2 的圆心为 (60+15,15)=(75,15)(60 + 15, 15) = (75, 15)。三角形 QUVQUV 与三角形 QPRQPR 相似，比例为 QUQP=3090=13\frac{QU}{QP} = \frac{30}{90} = \frac{1}{3}，所以其内切圆半径为 1010，C3C_3 的圆心为 (10,60+10)=(10,70)(10, 60 + 10) = (10, 70)。

两圆心距离的平方为 652+55265^2 + 55^2 =4225+3025= 4225 + 3025 =7250= 7250 =10⋅725= 10 \cdot 725，所以 n=725n = 725。

The right angle is at P,P, so place P=(0,0),P = (0, 0), Q=(0,90),Q = (0, 90), R=(120,0).R = (120, 0). The inradius of a right triangle is half the difference between the sum of the legs and the hypotenuse: r1=90+120−1502=30,r_1 = \frac{90 + 120 - 150}{2} = 30, so C1C_1 has center (30,30).(30, 30). The tangent line to C1C_1 perpendicular to PR‾\overline{PR} (on the side toward RR) is x=60,x = 60, and the tangent perpendicular to PQ‾\overline{PQ} (toward QQ) is y=60.y = 60.

Triangle RSTRST is similar to triangle RPQRPQ with ratio RSRP=60120=12,\frac{RS}{RP} = \frac{60}{120} = \frac{1}{2}, so its inradius is 1515 and its incircle C2C_2 is centered at (60+15,15)=(75,15).(60 + 15, 15) = (75, 15). Triangle QUVQUV is similar to triangle QPRQPR with ratio QUQP=3090=13,\frac{QU}{QP} = \frac{30}{90} = \frac{1}{3}, so its inradius is 1010 and C3C_3 is centered at (10,60+10)=(10,70).(10, 60 + 10) = (10, 70).

The squared distance is 652+55265^2 + 55^2 =4225+3025= 4225 + 3025 =7250= 7250 =10⋅725,= 10 \cdot 725, so n=725.n = 725.

8.

某函数 ff 满足：对所有正实数 xx，都有 f(3x)=3f(x)f(3x) = 3f(x)；并且当 1≤x≤31 \le x \le 3 时，f(x)=1−∣x−2∣f(x) = 1 - |x - 2|。求满足 f(x)=f(2001)f(x) = f(2001) 的最小 xx。

A certain function ff has the properties that f(3x)=3f(x)f(3x) = 3f(x) for all positive real values of x,x, and that f(x)=1−∣x−2∣f(x) = 1 - |x - 2| for 1≤x≤3.1 \le x \le 3. Find the smallest xx for which f(x)=f(2001).f(x) = f(2001).

答案：429
知识点：函数方程绝对值
难度评级：2560
小提示：

用 f(x)=36f(x36)f(x) = 3^6 f\left(\frac{x}{3^6}\right) 计算 f(2001)f(2001)，因为 2001729\frac{2001}{729} 落在 [1,3][1, 3] 中。

Use f(x)=36f(x36)f(x) = 3^6 f\left(\frac{x}{3^6}\right) to compute f(2001),f(2001), since 2001729\frac{2001}{729} lies in [1,3][1, 3]

大提示：

在 [3k,3k+1][3^k, 3^{k+1}] 上，图像是一个峰值为 3k3^k 的帐篷形；找第一个峰值达到 f(2001)f(2001) 的区间。

On [3k,3k+1][3^k, 3^{k+1}] the graph is a tent peaking at height 3k,3^k, so look in the first interval whose peak reaches f(2001)f(2001)

解答：

连续使用 f(3x)=3f(x)f(3x) = 3f(x) 六次得 f(2001)=36f(2001729)f(2001) = 3^6 f\left(\frac{2001}{729}\right)，且 2001729\frac{2001}{729} 在 [1,3][1, 3] 中，所以 f(2001)=729(1−∣2001729−2∣)=729−∣2001−1458∣=729−543=186。 \begin{aligned} f(2001) &= 729\left(1 - \left|\tfrac{2001}{729} - 2\right|\right) \\ &= 729 - |2001 - 1458| \\ &= 729 - 543 = 186 \end{aligned}\text{。}

对 x∈[3k,3k+1]x \in [3^k, 3^{k+1}] 有 f(x)=3kf(x3k)f(x) = 3^k f\left(\frac{x}{3^k}\right) =3k(1−∣x3k−2∣)= 3^k\left(1 - \left|\frac{x}{3^k} - 2\right|\right)，这是一个最大值为 3k3^k 的帐篷形函数。要取得 186186，需要 3k≥1863^k \ge 186，所以 k≥5k \ge 5。最小的解会在 [243,729][243, 729] 中，此时 f(x)=243−∣x−486∣f(x) = 243 - |x - 486|。

令 243−∣x−486∣=186243 - |x - 486| = 186，得 ∣x−486∣=57|x - 486| = 57，所以 x=429x = 429 或 x=543x = 543。最小的 xx 是 429429。

Applying f(3x)=3f(x)f(3x) = 3f(x) six times gives f(2001)=36f(2001729),f(2001) = 3^6 f\left(\frac{2001}{729}\right), and 2001729\frac{2001}{729} lies in [1,3],[1, 3], so f(2001)=729(1−∣2001729−2∣)=729−∣2001−1458∣=729−543=186. \begin{aligned} f(2001) &= 729\left(1 - \left|\tfrac{2001}{729} - 2\right|\right) \\ &= 729 - |2001 - 1458| \\ &= 729 - 543 = 186. \end{aligned}

For x∈[3k,3k+1],x \in [3^k, 3^{k+1}], we have f(x)=3kf(x3k)f(x) = 3^k f\left(\frac{x}{3^k}\right) =3k(1−∣x3k−2∣),= 3^k\left(1 - \left|\frac{x}{3^k} - 2\right|\right), a tent whose maximum value is 3k.3^k. To achieve 186186 we need 3k≥186,3^k \ge 186, so k≥5,k \ge 5, and the smallest solutions lie in [243,729],[243, 729], where f(x)=243−∣x−486∣.f(x) = 243 - |x - 486|.

Setting 243−∣x−486∣=186243 - |x - 486| = 186 gives ∣x−486∣=57,|x - 486| = 57, so x=429x = 429 or x=543.x = 543. The smallest xx is 429.429.

9.

一个 33 乘 33 的单位正方形网格中，每个单位正方形要被涂成蓝色或红色。对每个小方格，两种颜色被使用的概率相等。得到一个不含 22 乘 22 红色正方形的网格的概率为 mn\frac{m}{n}，其中 mm 和 nn 是互质正整数。求 m+nm + n。

Each unit square of a 33-by-33 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 22-by-22 red square is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

答案：929
知识点：容斥原理对立事件概率
难度评级：2710
小提示：

对四个可能的 22 乘 22 全红方块位置，用容斥法计算补事件。

Count the complement by inclusion-exclusion on the four possible positions of a 22-by-22 red square

大提示：

两个相交的 22 乘 22 方块覆盖 66 或 77 个格子，任意三个覆盖 88 个格子，四个全部覆盖 99 个格子；每个被强制为红色的格子贡献因子 12\frac{1}{2}。

Two overlapping 22-by-22 blocks cover 66 or 77 cells, any three cover 8,8, and all four cover 9;9; each forced cell contributes a factor 12\frac{1}{2}

解答：

对四个可能位置使用容斥法，计算网格中含有一个全红 22 乘 22 方块的概率。一个方块强制 44 个格子为红色；两个方块若共边，则强制 66 个格子为红色，共有 44 对；另外两对对角位置各强制 77 个格子为红色。任意三个方块强制 88 个格子，四个方块强制全部 99 个格子。

每种强制红色格子的配置概率为 (12)格子数\left(\frac{1}{2}\right)^{\text{格子数}}，所以至少有一个全红方块的概率是 4⋅116−(4⋅164+2⋅1128)+4⋅1256−1512=128−40+8−1512=95512。 \begin{aligned} &4 \cdot \frac{1}{16} - \left(4 \cdot \frac{1}{64} + 2 \cdot \frac{1}{128}\right) \\ &\quad {}+ 4 \cdot \frac{1}{256} - \frac{1}{512} \\ &= \frac{128 - 40 + 8 - 1}{512} = \frac{95}{512} \end{aligned}\text{。}

所求概率为 1−95512=4175121 - \frac{95}{512} = \frac{417}{512}，且 417=3⋅139417 = 3 \cdot 139 与 512512 互质，因此 m+n=417+512=929m + n = 417 + 512 = 929。

Compute the probability that the grid does contain an all-red 22-by-22 block by inclusion-exclusion over the four possible positions. One block forces 44 cells; two blocks sharing an edge force 66 cells (44 such pairs), while the two diagonal pairs force 7;7; any three blocks force 88 cells, and all four force all 9.9.

Each configuration of forced red cells has probability (12)cells,\left(\frac{1}{2}\right)^{\text{cells}}, so the probability of at least one red block is 4⋅116−(4⋅164+2⋅1128)+4⋅1256−1512=128−40+8−1512=95512. \begin{aligned} &4 \cdot \frac{1}{16} - \left(4 \cdot \frac{1}{64} + 2 \cdot \frac{1}{128}\right) \\ &\quad {}+ 4 \cdot \frac{1}{256} - \frac{1}{512} \\ &= \frac{128 - 40 + 8 - 1}{512} = \frac{95}{512}. \end{aligned}

The desired probability is 1−95512=417512,1 - \frac{95}{512} = \frac{417}{512}, and 417=3⋅139417 = 3 \cdot 139 is coprime to 512,512, so m+n=417+512=929.m + n = 417 + 512 = 929.

10.

有多少个 10011001 的正整数倍可以表示为 10j−10i10^{j} - 10^{i} 的形式，其中 ii 和 jj 是整数且 0≤i<j≤990 \le i \lt j \le 99？

How many positive integer multiples of 10011001 can be expressed in the form 10j−10i,10^{j} - 10^{i}, where ii and jj are integers and 0≤i<j≤99?0 \le i \lt j \le 99?

答案：784
知识点：乘法阶模运算数对计数
难度评级：2710
小提示：

分解 10j−10i=10i(10j−i−1)10^j - 10^i = 10^i(10^{j-i} - 1)，并注意 10011001 与 10i10^i 互质。

Factor 10j−10i=10i(10j−i−1)10^j - 10^i = 10^i(10^{j-i} - 1) and note that 10011001 is coprime to 10i10^i

大提示：

10k−110^k - 1 能被 10011001 整除当且仅当 kk 是 66 的倍数，所以只需数出满足 0≤i<j≤990 \le i \lt j \le 99 且 j−ij - i 为 66 的倍数的有序对。

10k−110^k - 1 is divisible by 10011001 exactly when kk is a multiple of 6,6, so count pairs 0≤i<j≤990 \le i \lt j \le 99 with j−ij - i a multiple of 66

解答：

分解得 10j−10i=10i(10j−i−1)10^j - 10^i = 10^i(10^{j-i} - 1)。由于 1001=7⋅11⋅131001 = 7 \cdot 11 \cdot 13 与 10i10^i 互质，我们需要 10j−i−110^{j-i} - 1 能被 10011001 整除。1010 模 77、模 1111、模 1313 的乘法阶分别为 66、22、66，所以 10k≡1(mod1001)10^k \equiv 1 \pmod{1001} 当且仅当 kk 是 66 的倍数。不同的 (i,j)(i, j) 给出不同的值，因此只需计数这些有序对。

若 j−i=6dj - i = 6d，其中 1≤d≤161 \le d \le 16，则 ii 可取 0,1,…,99−6d0, 1, \ldots, 99 - 6d，共有 100−6d100 - 6d 种。总数为 ∑d=116(100−6d)=94+88+⋯+4=(94+4)⋅162=784。 \begin{aligned} &\sum_{d=1}^{16} (100 - 6d) \\ &= 94 + 88 + \cdots + 4 \\ &= \frac{(94 + 4) \cdot 16}{2} \\ &= 784 \end{aligned}\text{。}

Factor 10j−10i=10i(10j−i−1).10^j - 10^i = 10^i(10^{j-i} - 1). Since 1001=7⋅11⋅131001 = 7 \cdot 11 \cdot 13 is coprime to 10i,10^i, we need 10j−i−110^{j-i} - 1 to be divisible by 1001.1001. The multiplicative order of 1010 is 66 modulo 7,7, 22 modulo 11,11, and 66 modulo 13,13, so 10k≡1(mod1001)10^k \equiv 1 \pmod{1001} exactly when kk is a multiple of 6.6. Distinct pairs (i,j)(i, j) give distinct values, so we just count the pairs.

For j−i=6dj - i = 6d with 1≤d≤16,1 \le d \le 16, the index ii can be 0,1,…,99−6d,0, 1, \ldots, 99 - 6d, giving 100−6d100 - 6d choices. The total is ∑d=116(100−6d)=94+88+⋯+4=(94+4)⋅162=784. \begin{aligned} &\sum_{d=1}^{16} (100 - 6d) \\ &= 94 + 88 + \cdots + 4 \\ &= \frac{(94 + 4) \cdot 16}{2} \\ &= 784. \end{aligned}

11.

Truncator 俱乐部与另外六支队同在一个足球联赛中，并与每支队比赛一次。在它的 66 场比赛中，Truncator 俱乐部获胜、失利、打平的概率各为 13\frac{1}{3}。Truncator 俱乐部赛季结束时胜场数多于负场数的概率为 mn\frac{m}{n}，其中 mm 和 nn 是互质正整数。求 m+nm + n。

Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 66 matches, the probabilities that Club Truncator will win, lose, or tie are each 13.\frac{1}{3}. The probability that Club Truncator will finish the season with more wins than losses is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

答案：341
知识点：基本概率多重集排列补集计数对称性
难度评级：2560
小提示：

由对称性，胜场数多于负场数和负场数多于胜场数的概率相等。

By symmetry, more wins than losses and more losses than wins are equally likely

大提示：

求胜场数等于负场数的概率：对每个 kk，数出恰有 kk 场胜、kk 场负、6−2k6 - 2k 场平的结果数。

Find the probability of equally many wins and losses: count outcomes with kk wins, kk losses, and 6−2k6 - 2k ties for each kk

解答：

交换胜负是保持概率不变的对称操作，所以胜场多于负场的概率 PP 等于负场多于胜场的概率。因此 P=1−p02P = \frac{1 - p_0}{2}，其中 p0p_0 是胜场数和负场数相等的概率。

若有 kk 场胜、kk 场负、6−2k6 - 2k 场平，则这些结果可排列成 6!k! k! (6−2k)!\frac{6!}{k! \, k! \, (6 - 2k)!} 种。当 k=0,1,2,3k = 0, 1, 2, 3 时，这些数量分别为 11、3030、9090、2020，总计 141141。所有 36=7293^6 = 729 个结果序列等可能，所以 p0=141729=47243p_0 = \frac{141}{729} = \frac{47}{243}。

因此 P=12(1−47243)=98243P = \frac{1}{2}\left(1 - \frac{47}{243}\right) = \frac{98}{243}，所以 m+n=98+243=341m + n = 98 + 243 = 341。

Swapping wins and losses is a probability-preserving symmetry, so the probability PP of more wins than losses equals the probability of more losses than wins, giving P=1−p02,P = \frac{1 - p_0}{2}, where p0p_0 is the probability of equally many wins and losses.

An outcome with kk wins, kk losses, and 6−2k6 - 2k ties can be arranged in 6!k! k! (6−2k)!\frac{6!}{k! \, k! \, (6 - 2k)!} ways: 1,1, 30,30, 90,90, 2020 for k=0,1,2,3,k = 0, 1, 2, 3, totaling 141.141. Each of the 36=7293^6 = 729 outcome sequences is equally likely, so p0=141729=47243.p_0 = \frac{141}{729} = \frac{47}{243}.

Therefore P=12(1−47243)=98243,P = \frac{1}{2}\left(1 - \frac{47}{243}\right) = \frac{98}{243}, and m+n=98+243=341.m + n = 98 + 243 = 341.

12.

对一个三角形，它的中点三角形由连接三边中点得到。递归定义一列多面体 Pi\mathcal{P}_i 如下：P0\mathcal{P}_0 是体积为 11 的正四面体。为了得到 Pi+1\mathcal{P}_{i+1}，将 Pi\mathcal{P}_i 的每个面上的中点三角形替换为一个向外突出的正四面体，该中点三角形作为这个正四面体的一个面。P3\mathcal{P}_3 的体积为 mn\frac{m}{n}，其中 mm 和 nn 是互质正整数。求 m+nm + n。

Given a triangle, its midpoint triangle is obtained by joining the midpoints of its sides. A sequence of polyhedra Pi\mathcal{P}_i is defined recursively as follows: P0\mathcal{P}_0 is a regular tetrahedron whose volume is 1.1. To obtain Pi+1,\mathcal{P}_{i+1}, replace the midpoint triangle of every face of Pi\mathcal{P}_i by an outward-pointing regular tetrahedron that has the midpoint triangle as a face. The volume of P3\mathcal{P}_3 is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

答案：101
知识点：体积长度、面积与体积的缩放关系递推等比数列
难度评级：2990
小提示：

Pi\mathcal{P}_i 的每个面会被替换成 66 个半边长的面，所以 Pi\mathcal{P}_i 有 4⋅6i4 \cdot 6^i 个面。

Each face of Pi\mathcal{P}_i is replaced by 66 half-size faces, so Pi\mathcal{P}_i has 4⋅6i4 \cdot 6^i faces

大提示：

第 i+1i + 1 步添加的每个四面体体积为 (18)i+1\left(\frac{1}{8}\right)^{i+1}，所以添加的总体积为 12(34)i\frac{1}{2}\left(\frac{3}{4}\right)^i。

The tetrahedra added at step i+1i + 1 each have volume (18)i+1,\left(\frac{1}{8}\right)^{i+1}, so the total volume added is 12(34)i\frac{1}{2}\left(\frac{3}{4}\right)^i

解答：

在某个面上的中点三角形处贴上一个四面体，会把原来的面替换成 66 个边长为一半的等边三角形：33 个角上的三角形，加上新四面体暴露出的 33 个面。因此 Pi\mathcal{P}_i 的所有面全等，边长为原来的 (12)i\left(\frac{1}{2}\right)^i 倍，且 Pi\mathcal{P}_i 有 4⋅6i4 \cdot 6^i 个面。

从 Pi\mathcal{P}_i 到 Pi+1\mathcal{P}_{i+1} 时，每个面上粘上一个正四面体；每个新四面体都与 P0\mathcal{P}_0 相似，比例为 (12)i+1\left(\frac{1}{2}\right)^{i+1}，因而体积为 (18)i+1\left(\frac{1}{8}\right)^{i+1}。增加的体积为 4⋅6i(18)i+1=12(34)i。4 \cdot 6^i \left(\frac{1}{8}\right)^{i+1} = \frac{1}{2}\left(\frac{3}{4}\right)^i\text{。}

因此 P3\mathcal{P}_3 的体积为 1+12+38+932=69321 + \frac{1}{2} + \frac{3}{8} + \frac{9}{32} = \frac{69}{32}，所以 m+n=69+32=101m + n = 69 + 32 = 101。

Attaching a tetrahedron over the midpoint triangle of a face replaces that face by 66 equilateral triangles of half the side length: the 33 corner triangles plus 33 exposed faces of the new tetrahedron. So all faces of Pi\mathcal{P}_i are congruent, with side (12)i\left(\frac{1}{2}\right)^i times the original, and Pi\mathcal{P}_i has 4⋅6i4 \cdot 6^i faces.

Passing from Pi\mathcal{P}_i to Pi+1\mathcal{P}_{i+1} glues one regular tetrahedron onto each face; each is similar to P0\mathcal{P}_0 with ratio (12)i+1,\left(\frac{1}{2}\right)^{i+1}, hence has volume (18)i+1.\left(\frac{1}{8}\right)^{i+1}. The volume added is 4⋅6i(18)i+1=12(34)i.4 \cdot 6^i \left(\frac{1}{8}\right)^{i+1} = \frac{1}{2}\left(\frac{3}{4}\right)^i.

Therefore the volume of P3\mathcal{P}_3 is 1+12+38+932=6932,1 + \frac{1}{2} + \frac{3}{8} + \frac{9}{32} = \frac{69}{32}, and m+n=69+32=101.m + n = 69 + 32 = 101.

13.

在四边形 ABCDABCD 中，∠BAD≅∠ADC\angle BAD \cong \angle ADC 且 ∠ABD≅∠BCD\angle ABD \cong \angle BCD，并且 AB=8AB = 8、BD=10BD = 10、BC=6BC = 6。长度 CDCD 可写成 mn\frac{m}{n} 的形式，其中 mm 和 nn 是互质正整数。求 m+nm + n。

In quadrilateral ABCD,ABCD, ∠BAD≅∠ADC\angle BAD \cong \angle ADC and ∠ABD≅∠BCD,\angle ABD \cong \angle BCD, AB=8,AB = 8, BD=10,BD = 10, and BC=6.BC = 6. The length CDCD may be written in the form mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

答案：69
知识点：相似等腰三角形导角
难度评级：2990
小提示：

延长 AB‾\overline{AB} 和 DC‾\overline{DC} 交于 PP；AA 与 DD 处的相等角会迫使 PA=PDPA = PD。

Extend AB‾\overline{AB} and DC‾\overline{DC} to meet at P;P; the equal angles at AA and DD force PA=PDPA = PD

大提示：

角 PBDPBD 和 PCBPCB 分别是题中相等角的补角，所以三角形 PCBPCB 与 PBDPBD 相似。

Angles PBDPBD and PCBPCB are supplements of the given equal angles, so triangles PCBPCB and PBDPBD are similar

解答：

将 AB‾\overline{AB} 越过 BB 延长，将 DC‾\overline{DC} 越过 CC 延长，交于 PP。因为 ∠PAD=∠PDA\angle PAD = \angle PDA，三角形 APDAPD 是等腰三角形，且 PA=PDPA = PD。另外，∠PBD=180∘−∠ABD\angle PBD = 180^\circ - \angle ABD，∠PCB=180∘−∠BCD\angle PCB = 180^\circ - \angle BCD，所以 ∠PBD=∠PCB\angle PBD = \angle PCB。

三角形 PCBPCB 与 PBDPBD 共用角 PP，并且 ∠PCB=∠PBD\angle PCB = \angle PBD，所以它们相似，得到 PCPB=PBPD=CBBD=35。\frac{PC}{PB} = \frac{PB}{PD} = \frac{CB}{BD} = \frac{3}{5}\text{。}因为 PB=PA−8=PD−8PB = PA - 8 = PD - 8，中间的比例为 PD−8PD=35\frac{PD - 8}{PD} = \frac{3}{5}，所以 PD=20PD = 20，且 PB=12PB = 12。于是 PC=35⋅12=365PC = \frac{3}{5} \cdot 12 = \frac{36}{5}。

最后 CD=PD−PCCD = PD - PC =20−365= 20 - \frac{36}{5} =645= \frac{64}{5}，已经是最简形式，所以 m+n=64+5=69m + n = 64 + 5 = 69。

Extend AB‾\overline{AB} beyond BB and DC‾\overline{DC} beyond CC to meet at P.P. Since ∠PAD=∠PDA,\angle PAD = \angle PDA, triangle APDAPD is isosceles with PA=PD.PA = PD. Also ∠PBD=180∘−∠ABD\angle PBD = 180^\circ - \angle ABD and ∠PCB=180∘−∠BCD,\angle PCB = 180^\circ - \angle BCD, so ∠PBD=∠PCB.\angle PBD = \angle PCB.

Triangles PCBPCB and PBDPBD share angle PP and have ∠PCB=∠PBD,\angle PCB = \angle PBD, so they are similar, giving PCPB=PBPD=CBBD=35.\frac{PC}{PB} = \frac{PB}{PD} = \frac{CB}{BD} = \frac{3}{5}. Since PB=PA−8=PD−8,PB = PA - 8 = PD - 8, the middle ratio reads PD−8PD=35,\frac{PD - 8}{PD} = \frac{3}{5}, so PD=20PD = 20 and PB=12.PB = 12. Then PC=35⋅12=365.PC = \frac{3}{5} \cdot 12 = \frac{36}{5}.

Finally CD=PD−PCCD = PD - PC =20−365= 20 - \frac{36}{5} =645,= \frac{64}{5}, which is in lowest terms, so m+n=64+5=69.m + n = 64 + 5 = 69.

14.

有 2n2n 个复数同时满足 z28−z8−1=0z^{28} - z^{8} - 1 = 0 和 ∣z∣=1|z| = 1。这些数形如 zm=cos⁡θm+isin⁡θmz_{m} = \cos\theta_{m} + i\sin\theta_{m}，其中 0≤θ1<θ2<…<θ2n<3600 \le \theta_{1} \lt \theta_{2} \lt \ldots \lt \theta_{2n} \lt 360，且角以度为单位。求 θ2+θ4+⋯+θ2n\theta_{2} + \theta_{4} + \cdots + \theta_{2n} 的值。

There are 2n2n complex numbers that satisfy both z28−z8−1=0z^{28} - z^{8} - 1 = 0 and ∣z∣=1.|z| = 1. These numbers have the form zm=cos⁡θm+isin⁡θm,z_{m} = \cos\theta_{m} + i\sin\theta_{m}, where 0≤θ1<θ2<…<θ2n<3600 \le \theta_{1} \lt \theta_{2} \lt \ldots \lt \theta_{2n} \lt 360 and angles are measured in degrees. Find the value of θ2+θ4+⋯+θ2n.\theta_{2} + \theta_{4} + \cdots + \theta_{2n}.

答案：840
知识点：复数棣莫弗定理模运算
难度评级：3060
小提示：

将方程改写为 z8(z20−1)=1z^8(z^{20} - 1) = 1，并取绝对值：∣z20−1∣=1|z^{20} - 1| = 1。

Rewrite the equation as z8(z20−1)=1z^8(z^{20} - 1) = 1 and take absolute values: ∣z20−1∣=1|z^{20} - 1| = 1

大提示：

z20z^{20} 到 11 和 00 的距离都为 11，所以 z20=cos⁡60∘±isin⁡60∘z^{20} = \cos 60^\circ \pm i\sin 60^\circ；再推出 z4z^4，从而推出 θ\theta 模 9090。

z20z^{20} is at distance 11 from both 00 and 1,1, so z20=cos⁡60∘±isin⁡60∘;z^{20} = \cos 60^\circ \pm i\sin 60^\circ; deduce z4z^4 and hence θ\theta mod 9090

解答：

记 cis⁡θ=cos⁡θ+isin⁡θ\operatorname{cis}\theta = \cos\theta + i\sin\theta。原方程为 z8(z20−1)=1z^8(z^{20} - 1) = 1。取绝对值并使用 ∣z∣=1|z| = 1，得 ∣z20−1∣=1|z^{20} - 1| = 1，所以 z20z^{20} 到 11 和 00 的距离都为 11：它等于 cis⁡(±60∘)\operatorname{cis}(\pm 60^\circ)。

若 z20=cis⁡60∘z^{20} = \operatorname{cis} 60^\circ，则 z20−1=cis⁡120∘z^{20} - 1 = \operatorname{cis} 120^\circ，所以 z8=cis⁡(−120∘)z^8 = \operatorname{cis}(-120^\circ)，并且 z4=z20(z8)2=cis⁡(60∘+240∘)=cis⁡300∘， \begin{aligned} z^4 &= \frac{z^{20}}{(z^8)^2} \\ &= \operatorname{cis}(60^\circ + 240^\circ) = \operatorname{cis} 300^\circ \end{aligned}\text{，}因此 4θ≡300∘4\theta \equiv 300^\circ，即 θ≡75∘(mod90∘)\theta \equiv 75^\circ \pmod{90^\circ}。反过来，每个这样的 θ\theta 都满足条件，因为此时 z20=(z4)5=cis⁡60∘z^{20} = (z^4)^5 = \operatorname{cis} 60^\circ，且 z8=(z4)2=cis⁡(−120∘)z^8 = (z^4)^2 = \operatorname{cis}(-120^\circ)。类似地，z20=cis⁡(−60∘)z^{20} = \operatorname{cis}(-60^\circ) 的情形恰好给出 θ≡15∘(mod90∘)\theta \equiv 15^\circ \pmod{90^\circ}。

因此 2n=82n = 8 个角按从小到大排列为 15,75,105,165,15, 75, 105, 165, 195,255,285,345195, 255, 285, 345，并且 θ2+θ4+θ6+θ8\theta_2 + \theta_4 + \theta_6 + \theta_8 =75+165+255+345= 75 + 165 + 255 + 345 =840= 840。

Write cis⁡θ=cos⁡θ+isin⁡θ.\operatorname{cis}\theta = \cos\theta + i\sin\theta. The equation says z8(z20−1)=1.z^8(z^{20} - 1) = 1. Taking absolute values and using ∣z∣=1|z| = 1 gives ∣z20−1∣=1,|z^{20} - 1| = 1, so z20z^{20} is at distance 11 from both 00 and 1:1: it is cis⁡(±60∘).\operatorname{cis}(\pm 60^\circ).

If z20=cis⁡60∘,z^{20} = \operatorname{cis} 60^\circ, then z20−1=cis⁡120∘,z^{20} - 1 = \operatorname{cis} 120^\circ, so z8=cis⁡(−120∘)z^8 = \operatorname{cis}(-120^\circ) and z4=z20(z8)2=cis⁡(60∘+240∘)=cis⁡300∘, \begin{aligned} z^4 &= \frac{z^{20}}{(z^8)^2} \\ &= \operatorname{cis}(60^\circ + 240^\circ) = \operatorname{cis} 300^\circ, \end{aligned} which means 4θ≡300∘,4\theta \equiv 300^\circ, i.e. θ≡75∘(mod90∘).\theta \equiv 75^\circ \pmod{90^\circ}. Conversely every such θ\theta works, since then z20=(z4)5=cis⁡60∘z^{20} = (z^4)^5 = \operatorname{cis} 60^\circ and z8=(z4)2=cis⁡(−120∘).z^8 = (z^4)^2 = \operatorname{cis}(-120^\circ). The case z20=cis⁡(−60∘)z^{20} = \operatorname{cis}(-60^\circ) similarly gives exactly θ≡15∘(mod90∘).\theta \equiv 15^\circ \pmod{90^\circ}.

So the 2n=82n = 8 angles in increasing order are 15,75,105,165,15, 75, 105, 165, 195,255,285,345,195, 255, 285, 345, and θ2+θ4+θ6+θ8\theta_2 + \theta_4 + \theta_6 + \theta_8 =75+165+255+345= 75 + 165 + 255 + 345 =840.= 840.

15.

设 EFGHEFGH、EFDCEFDC 和 EHBCEHBC 是一个立方体的三个相邻正方形面，且 EC=8EC = 8，AA 是该立方体的第八个顶点。令 II、JJ、KK 分别在 EF‾\overline{EF}、EH‾\overline{EH}、EC‾\overline{EC} 上，满足 EI=EJ=EK=2EI = EJ = EK = 2。在立方体中钻出一个隧道后，得到立体 SS。隧道的侧壁是平行于 AE‾\overline{AE} 的平面，并且分别包含边 IJ‾\overline{IJ}、JK‾\overline{JK} 和 KI‾\overline{KI}。SS 的表面积（包括隧道壁）为 m+npm + n\sqrt{p}，其中 mm、nn、pp 是正整数，且 pp 不被任何素数的平方整除。求 m+n+pm + n + p。

Let EFGH,EFGH, EFDC,EFDC, and EHBCEHBC be three adjacent square faces of a cube, for which EC=8,EC = 8, and let AA be the eighth vertex of the cube. Let I,I, J,J, and KK be points on EF‾,\overline{EF}, EH‾,\overline{EH}, and EC‾,\overline{EC}, respectively, so that EI=EJ=EK=2.EI = EJ = EK = 2. A solid SS is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to AE‾,\overline{AE}, and containing the edges IJ‾,\overline{IJ}, JK‾,\overline{JK}, and KI‾.\overline{KI}. The surface area of S,S, including the walls of the tunnel, is m+np,m + n\sqrt{p}, where m,m, n,n, and pp are positive integers and pp is not divisible by the square of any prime. Find m+n+p.m + n + p.

答案：417
知识点：正方体坐标几何表面积
难度评级：3370
小提示：

取 A=(0,0,0)A = (0,0,0)、E=(8,8,8)E = (8,8,8)；每个隧道壁都是过三角形 IJKIJK 一条边且方向为 (1,1,1)(1,1,1) 的平面。

Take A=(0,0,0)A = (0,0,0) and E=(8,8,8);E = (8,8,8); each tunnel wall is a plane through one edge of triangle IJKIJK in the direction (1,1,1)(1,1,1)

大提示：

隧道从六个立方体面各切去面积 22，且三面隧道壁每面都是一个 22×632\sqrt{2} \times 6\sqrt{3} 的长方形加一个小三角形。

The tunnel cuts area 22 from each of the six cube faces, and each of the three walls is a 22×632\sqrt{2} \times 6\sqrt{3} rectangle plus a small triangle

解答：

令 A=(0,0,0)A = (0,0,0)、E=(8,8,8)E = (8,8,8)，于是 I=(6,8,8)I = (6,8,8)、J=(8,6,8)J = (8,6,8)、K=(8,8,6)K = (8,8,6)，且 AE‾\overline{AE} 的方向为 (1,1,1)(1,1,1)。从 II 沿这个方向的直线在 L=(0,2,2)L = (0,2,2) 处离开立方体；类似地，JJ 和 KK 分别对应 M=(2,0,2)M = (2,0,2) 和 N=(2,2,0)N = (2,2,0)。经过 II 和 JJ 的隧道壁所在平面为 2z=x+y+22z = x + y + 2，它也包含 LL 和 MM，并在 zz-轴上交于 O=(0,0,1)O = (0,0,1)；另外两面隧道壁对称地在 yy-轴和 xx-轴上分别交于 (0,1,0)(0,1,0) 和 (1,0,0)(1,0,0)。

现在求总表面积。靠近 EE 的三个立方体面各少掉一个直角边为 22 的直角三角形（如 IEJIEJ），剩余面积为 64−2=6264 - 2 = 62。靠近 AA 的三个面各少掉面积为 22 的四边形；在平面 z=0z = 0 上，这个四边形的顶点为 (0,0,0)(0,0,0)、(1,0,0)(1,0,0)、(2,2,0)(2,2,0)、(0,1,0)(0,1,0)。每面隧道壁都是像 ILOMJILOMJ 这样的五边形：长方形 ILMJILMJ 中，IJ=22IJ = 2\sqrt{2}、IL=63IL = 6\sqrt{3}，面积为 12612\sqrt{6}；等腰三角形 LOMLOM 的底边 LM=22LM = 2\sqrt{2}，高为 3\sqrt{3}，再增加 6\sqrt{6}，所以每面隧道壁面积为 13613\sqrt{6}。

总表面积为 6⋅62+3⋅136=372+3966 \cdot 62 + 3 \cdot 13\sqrt{6} = 372 + 39\sqrt{6}，所以 m+n+p=372+39+6m + n + p = 372 + 39 + 6 =417= 417。

Place A=(0,0,0)A = (0,0,0) and E=(8,8,8),E = (8,8,8), so that I=(6,8,8),I = (6,8,8), J=(8,6,8),J = (8,6,8), K=(8,8,6),K = (8,8,6), and AE‾\overline{AE} has direction (1,1,1).(1,1,1). The line through II in that direction leaves the cube at L=(0,2,2);L = (0,2,2); similarly JJ and KK lead to M=(2,0,2)M = (2,0,2) and N=(2,2,0).N = (2,2,0). The tunnel wall through II and JJ is the plane 2z=x+y+2,2z = x + y + 2, which also contains LL and MM and crosses the zz-axis at O=(0,0,1);O = (0,0,1); the other two walls behave symmetrically, crossing the yy- and xx-axes at (0,1,0)(0,1,0) and (1,0,0).(1,0,0).

Now add up the surface. Each of the three cube faces at EE loses a right triangle with legs 22 (such as IEJIEJ), leaving area 64−2=62.64 - 2 = 62. Each of the three faces at AA loses a quadrilateral of area 2:2: on the face z=0z = 0 its vertices are (0,0,0),(0,0,0), (1,0,0),(1,0,0), (2,2,0),(2,2,0), (0,1,0).(0,1,0). Each tunnel wall is a pentagon like ILOMJ:ILOMJ: the rectangle ILMJILMJ with IJ=22IJ = 2\sqrt{2} and IL=63IL = 6\sqrt{3} has area 126,12\sqrt{6}, and the isosceles triangle LOMLOM with base LM=22LM = 2\sqrt{2} and height 3\sqrt{3} adds 6,\sqrt{6}, for 13613\sqrt{6} per wall.

The total surface area is 6⋅62+3⋅136=372+396,6 \cdot 62 + 3 \cdot 13\sqrt{6} = 372 + 39\sqrt{6}, so m+n+p=372+39+6m + n + p = 372 + 39 + 6 =417.= 417.