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2001 AIME I 详解

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所有题目均经美国数学协会（MAA）官方合法授权使用。

1.

求所有能被自身每一位数字整除的正两位整数之和。

Find the sum of all positive two-digit integers that are divisible by each of their digits.

知识点：数字整除性分类讨论
难度评级：1950
小提示：

若这个数为 10a+b10a + b，则能被十位数字 aa 整除会迫使 bb 是 aa 的倍数。

If the number is 10a+b,10a + b, divisibility by the tens digit aa forces bb to be a multiple of aa

大提示：

写成 b=kab = ka。再由 10a+b10a + b 能被 bb 整除，可知 k=1k = 1、22 或 55。

Write b=ka.b = ka. Then 10a+b10a + b divisible by bb forces k=1,k = 1, 2,2, or 55

解答：

设这个数为 10a+b10a + b，其中 aa 是十位数字，bb 是个位数字。因为 10a+b10a + b 能被 aa 整除，必须有 bb 能被 aa 整除，所以 b=kab = ka，其中 kk 为正整数。又因为 10a+b10a + b 能被 bb 整除，必须有 10a10a 能被 bb 整除，即 10a10a 能被 kaka 整除，所以 1010 能被 kk 整除。由于 b=ka≤9b = ka \le 9，可能的取值是 k=1k = 1、22 和 55。

当 k=1k = 1 时，数为 11,22,…,9911, 22, \ldots, 99，和为 11⋅45=49511 \cdot 45 = 495。当 k=2k = 2 时，数为 12,24,36,4812, 24, 36, 48，和为 120120。当 k=5k = 5 时，唯一的数是 1515。

总和为 495+120+15=630495 + 120 + 15 = 630。

Let the number be 10a+b10a + b with tens digit aa and units digit b.b. Since 10a+b10a + b is divisible by a,a, we need bb to be divisible by a,a, so b=kab = ka for some positive integer k.k. Since 10a+b10a + b is divisible by b,b, we need 10a10a to be divisible by b,b, that is 10a10a is divisible by ka,ka, so 1010 is divisible by k.k. Because b=ka≤9,b = ka \le 9, only k=1,k = 1, 2,2, and 55 are possible.

For k=1k = 1 the numbers are 11,22,…,99,11, 22, \ldots, 99, with sum 11⋅45=495.11 \cdot 45 = 495. For k=2k = 2 they are 12,24,36,48,12, 24, 36, 48, with sum 120.120. For k=5k = 5 the only one is 15.15.

The total is 495+120+15=630.495 + 120 + 15 = 630.

2.

一个由互不相同实数组成的有限集合 S\mathcal{S} 满足如下性质：S∪{1}\mathcal{S} \cup \{1\} 的平均数比 S\mathcal{S} 的平均数小 1313，而 S∪{2001}\mathcal{S} \cup \{2001\} 的平均数比 S\mathcal{S} 的平均数大 2727。求 S\mathcal{S} 的平均数。

A finite set S\mathcal{S} of distinct real numbers has the following properties: the mean of S∪{1}\mathcal{S} \cup \{1\} is 1313 less than the mean of S,\mathcal{S}, and the mean of S∪{2001}\mathcal{S} \cup \{2001\} is 2727 more than the mean of S.\mathcal{S}. Find the mean of S.\mathcal{S}.

知识点：平均数方程组
难度评级：1840
小提示：

设 S\mathcal{S} 有 nn 个元素，平均数为 xx，并把两个条件都写成方程。

Let S\mathcal{S} have nn elements with mean x,x, and translate each condition into an equation

大提示：

条件给出 x−13(n+1)=1x - 13(n+1) = 1 和 x+27(n+1)=2001x + 27(n+1) = 2001。将两式相减。

The conditions give x−13(n+1)=1x - 13(n+1) = 1 and x+27(n+1)=2001.x + 27(n+1) = 2001. Subtract them.

解答：

设 S\mathcal{S} 有 nn 个元素，平均数为 xx，则元素总和为 nxnx。两个条件分别为 nx+1n+1=x−13\frac{nx + 1}{n + 1} = x - 13 和 nx+2001n+1=x+27。\frac{nx + 2001}{n + 1} = x + 27\text{。}

清除分母，nx+1=(n+1)x−13(n+1)nx + 1 = (n+1)x - 13(n+1) 给出 x−13(n+1)=1x - 13(n+1) = 1，而 nx+2001nx + 2001 =(n+1)x+27(n+1)= (n+1)x + 27(n+1) 给出 x+27(n+1)=2001x + 27(n+1) = 2001。第二式减第一式得 40(n+1)=200040(n+1) = 2000，所以 n+1=50n + 1 = 50。

因此 x=1+13⋅50=651x = 1 + 13 \cdot 50 = 651。

Let S\mathcal{S} have nn elements with mean x,x, so the elements sum to nx.nx. The two conditions say nx+1n+1=x−13\frac{nx + 1}{n + 1} = x - 13 and nx+2001n+1=x+27.\frac{nx + 2001}{n + 1} = x + 27.

Clearing denominators, nx+1=(n+1)x−13(n+1)nx + 1 = (n+1)x - 13(n+1) gives x−13(n+1)=1,x - 13(n+1) = 1, and nx+2001nx + 2001 =(n+1)x+27(n+1)= (n+1)x + 27(n+1) gives x+27(n+1)=2001.x + 27(n+1) = 2001. Subtracting the first equation from the second yields 40(n+1)=2000,40(n+1) = 2000, so n+1=50.n + 1 = 50.

Then x=1+13⋅50=651.x = 1 + 13 \cdot 50 = 651.

3.

求方程 x2001+(12−x)2001=0x^{2001} + \left(\tfrac{1}{2} - x\right)^{2001} = 0 的所有根（实根和非实根）之和。已知该方程没有重根。

Find the sum of the roots, real and non-real, of the equation x2001+(12−x)2001=0,x^{2001} + \left(\tfrac{1}{2} - x\right)^{2001} = 0, given that there are no multiple roots.

知识点：多项式韦达定理二项式定理
难度评级：2300
小提示：

展开 (12−x)2001\left(\frac{1}{2} - x\right)^{2001} 时，x2001x^{2001} 项会被抵消，留下一个 20002000 次多项式。

Expanding (12−x)2001\left(\frac{1}{2} - x\right)^{2001} cancels the x2001x^{2001} term, leaving a polynomial of degree 20002000

大提示：

根之和等于 x1999x^{1999} 项系数的相反数除以 x2000x^{2000} 项系数。

The sum of the roots is minus the coefficient of x1999x^{1999} divided by the coefficient of x2000x^{2000}

解答：

用二项式定理展开 (12−x)2001\left(\frac{1}{2} - x\right)^{2001}。它的最高次项 (−x)2001=−x2001(-x)^{2001} = -x^{2001} 会与方程中的 x2001x^{2001} 抵消，所以剩下的是一个 20002000 次多项式：2001⋅12 x2000−(20012)14 x1999+⋯=0。 \begin{aligned} &2001 \cdot \frac{1}{2}\,x^{2000} \\ &\quad {}- \binom{2001}{2}\frac{1}{4}\,x^{1999} + \cdots = 0 \end{aligned}\text{。}

由韦达定理，20002000 个根的和为 (20012)420012=2001⋅2000820012=20004=500。 \begin{aligned} \frac{\frac{\binom{2001}{2}}{4}}{\frac{2001}{2}} &= \frac{\frac{2001 \cdot 2000}{8}}{\frac{2001}{2}} \\ &= \frac{2000}{4} = 500 \end{aligned}\text{。}

Expand (12−x)2001\left(\frac{1}{2} - x\right)^{2001} by the binomial theorem. Its leading term (−x)2001=−x2001(-x)^{2001} = -x^{2001} cancels the x2001x^{2001} in the equation, so what remains is a polynomial of degree 2000:2000: 2001⋅12 x2000−(20012)14 x1999+⋯=0. \begin{aligned} &2001 \cdot \frac{1}{2}\,x^{2000} \\ &\quad {}- \binom{2001}{2}\frac{1}{4}\,x^{1999} + \cdots = 0. \end{aligned}

By Vieta’s formulas, the sum of the 20002000 roots is (20012)420012=2001⋅2000820012=20004=500. \begin{aligned} \frac{\frac{\binom{2001}{2}}{4}}{\frac{2001}{2}} &= \frac{\frac{2001 \cdot 2000}{8}}{\frac{2001}{2}} \\ &= \frac{2000}{4} = 500. \end{aligned}

4.

在三角形 ABCABC 中，角 AA 和角 BB 分别为 6060 度和 4545 度。角 AA 的角平分线交 BC‾\overline{BC} 于 TT，且 AT=24AT = 24。三角形 ABCABC 的面积可写成 a+bca + b\sqrt{c}，其中 aa、bb、cc 是正整数，且 cc 不被任何质数的平方整除。求 a+b+ca + b + c。

In triangle ABC,ABC, angles AA and BB measure 6060 degrees and 4545 degrees, respectively. The bisector of angle AA intersects BC‾\overline{BC} at T,T, and AT=24.AT = 24. The area of triangle ABCABC can be written in the form a+bc,a + b\sqrt{c}, where a,a, b,b, and cc are positive integers, and cc is not divisible by the square of any prime. Find a+b+c.a + b + c.

知识点：导角等腰三角形特殊直角三角形三角形面积
难度评级：2390
小提示：

∠C=75∘\angle C = 75^\circ，而 ∠ATC=75∘\angle ATC = 75^\circ，所以 AC=AT=24AC = AT = 24。

∠C=75∘\angle C = 75^\circ and ∠ATC=75∘\angle ATC = 75^\circ too, so AC=AT=24AC = AT = 24

大提示：

从 CC 向 AB‾\overline{AB} 作高：它把 ABCABC 分成一个 3030-6060-9090 三角形和一个 4545-4545-9090 三角形。

Drop the altitude from CC to AB‾:\overline{AB}: it cuts ABCABC into a 3030-6060-9090 triangle and a 4545-4545-9090 triangle

解答：

因为 ∠A=60∘\angle A = 60^\circ 且 ∠B=45∘\angle B = 45^\circ，所以 ∠C=75∘\angle C = 75^\circ。在三角形 ATCATC 中，角 TAC=30∘TAC = 30^\circ（角 AA 的一半），所以 ∠ATC=180∘−30∘−75∘\angle ATC = 180^\circ - 30^\circ - 75^\circ =75∘= 75^\circ。因此三角形 ACTACT 为等腰三角形，且 AC=AT=24AC = AT = 24。

向 AB‾\overline{AB} 作高 CHCH。三角形 ACHACH 是 3030-6060-9090 三角形，所以 AH=12AH = 12，CH=123CH = 12\sqrt{3}。三角形 BCHBCH 是 4545-4545-9090 三角形，所以 BH=CH=123BH = CH = 12\sqrt{3}。

面积为 12⋅CH⋅AB\frac{1}{2} \cdot CH \cdot AB =12⋅123 (12+123)= \frac{1}{2} \cdot 12\sqrt{3}\,(12 + 12\sqrt{3}) =216+723= 216 + 72\sqrt{3}。因此 a+b+c=216+72+3=291a + b + c = 216 + 72 + 3 = 291。

Since ∠A=60∘\angle A = 60^\circ and ∠B=45∘,\angle B = 45^\circ, we have ∠C=75∘.\angle C = 75^\circ. In triangle ATC,ATC, angle TAC=30∘TAC = 30^\circ (half of angle AA), so ∠ATC=180∘−30∘−75∘\angle ATC = 180^\circ - 30^\circ - 75^\circ =75∘.= 75^\circ. Thus triangle ACTACT is isosceles with AC=AT=24.AC = AT = 24.

Drop the altitude CHCH to AB‾.\overline{AB}. Triangle ACHACH is 3030-6060-90,90, so AH=12AH = 12 and CH=123.CH = 12\sqrt{3}. Triangle BCHBCH is 4545-4545-90,90, so BH=CH=123.BH = CH = 12\sqrt{3}.

The area is 12⋅CH⋅AB\frac{1}{2} \cdot CH \cdot AB =12⋅123 (12+123)= \frac{1}{2} \cdot 12\sqrt{3}\,(12 + 12\sqrt{3}) =216+723.= 216 + 72\sqrt{3}. Then a+b+c=216+72+3=291.a + b + c = 216 + 72 + 3 = 291.

5.

一个等边三角形内接于椭圆 x2+4y2=4x^2 + 4y^2 = 4。三角形的一个顶点为 (0,1)(0, 1)，且有一条高在 yy-轴上。若每条边的长度为 mn\sqrt{\frac{m}{n}}，其中 mm 和 nn 是互质的正整数，求 m+nm + n。

An equilateral triangle is inscribed in the ellipse whose equation is x2+4y2=4.x^2 + 4y^2 = 4. One vertex of the triangle is (0,1),(0, 1), one altitude is contained in the yy-axis, and the length of each side is mn,\sqrt{\frac{m}{n}}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

知识点：椭圆等边三角形坐标几何对称性
难度评级：2510
小提示：

由对称性，另外两个顶点为 (±x,y)(\pm x, y)，且从 (0,1)(0, 1) 出发的边与正 xx-轴成 120∘120^\circ 角。

By symmetry the other two vertices are (±x,y),(\pm x, y), and the side through (0,1)(0, 1) makes a 120∘120^\circ angle with the positive xx-axis

大提示：

将 y=−3 x+1y = -\sqrt{3}\,x + 1 代入 x2+4y2=4x^2 + 4y^2 = 4 求出 xx；边长为 2x2x。

Substitute y=−3 x+1y = -\sqrt{3}\,x + 1 into x2+4y2=4x^2 + 4y^2 = 4 and solve for x;x; the side length is 2x2x

解答：

因为有一条高在 yy-轴上，另外两个顶点对称，可写为 (x,y)(x, y) 和 (−x,y)(-x, y)，其中 x>0x \gt 0。从 (0,1)(0,1) 到 (x,y)(x,y) 的边与正 xx-轴成 120∘120^\circ 角，所以它在直线 y=−3 x+1y = -\sqrt{3}\,x + 1 上。

代入椭圆方程 x2+4y2=4x^2 + 4y^2 = 4，得 x2+4(1−3x)2=4x^2 + 4(1 - \sqrt{3}x)^2 = 4，化简为 13x2−83 x=013x^2 - 8\sqrt{3}\,x = 0，因此 x=8313x = \frac{8\sqrt{3}}{13}。

边长为 2x=163132x = \frac{16\sqrt{3}}{13}，其平方为 768169\frac{768}{169}。由于 gcd⁡(768,169)=1\gcd(768, 169) = 1，答案为 768+169=937768 + 169 = 937。

Since one altitude lies along the yy-axis, the other two vertices are symmetric: (x,y)(x, y) and (−x,y)(-x, y) with x>0.x \gt 0. The side from (0,1)(0,1) to (x,y)(x,y) makes a 120∘120^\circ angle with the positive xx-axis, so it lies on the line y=−3 x+1.y = -\sqrt{3}\,x + 1.

Substituting into x2+4y2=4x^2 + 4y^2 = 4 gives x2+4(1−3x)2=4,x^2 + 4(1 - \sqrt{3}x)^2 = 4, which simplifies to 13x2−83 x=0,13x^2 - 8\sqrt{3}\,x = 0, so x=8313.x = \frac{8\sqrt{3}}{13}.

The side length is 2x=16313,2x = \frac{16\sqrt{3}}{13}, whose square is 768169.\frac{768}{169}. Since gcd⁡(768,169)=1,\gcd(768, 169) = 1, the answer is 768+169=937.768 + 169 = 937.

6.

一枚公平骰子掷四次。最后三次中的每一次点数都至少与前一次一样大的概率可写成 mn\frac{m}{n}，其中 mm 和 nn 是互质正整数。求 m+nm + n。

A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

知识点：基本概率隔板法双射
难度评级：2230
小提示：

每一组四个点数（允许重复）都恰好能按一种方式排列成非递减顺序。

Every collection of four values (with repeats allowed) can be arranged in exactly one non-decreasing order

大提示：

用隔板法计算从 {1,…,6}\{1, \ldots, 6\} 中选 44 个元素的多重集合数，结果为 (94)\binom{9}{4} 种。

Count multisets of size 44 from {1,…,6}\{1, \ldots, 6\} by stars and bars: (94)\binom{9}{4}

解答：

四次掷骰结果必须形成非递减序列。从 {1,…,6}\{1, \ldots, 6\} 中选出的任意四个数的多重集合，都恰好有一种非递减排列，所以成功结果数等于这类多重集合的数量。由隔板法（四个星和 55 个隔板），数量为 (94)=126\binom{9}{4} = 126。

概率为 12664=1261296=772\frac{126}{6^4} = \frac{126}{1296} = \frac{7}{72}，所以 m+n=7+72=79m + n = 7 + 72 = 79。

The rolls must form a non-decreasing sequence. Every multiset of four values from {1,…,6}\{1, \ldots, 6\} can be arranged in non-decreasing order in exactly one way, so the number of successful outcomes equals the number of such multisets. By stars and bars (4 stars and 55 dividers), that count is (94)=126.\binom{9}{4} = 126.

The probability is 12664=1261296=772,\frac{126}{6^4} = \frac{126}{1296} = \frac{7}{72}, so m+n=7+72=79.m + n = 7 + 72 = 79.

7.

三角形 ABCABC 满足 AB=21AB = 21、AC=22AC = 22、BC=20BC = 20。点 DD 和 EE 分别在 AB‾\overline{AB} 和 AC‾\overline{AC} 上，DE‾\overline{DE} 平行于 BC‾\overline{BC}，且经过三角形 ABCABC 的内心。若 DE=mnDE = \frac{m}{n}，其中 mm 和 nn 是互质的正整数，求 m+nm + n。

Triangle ABCABC has AB=21,AB = 21, AC=22,AC = 22, and BC=20.BC = 20. Points DD and EE are located on AB‾\overline{AB} and AC‾,\overline{AC}, respectively, such that DE‾\overline{DE} is parallel to BC‾\overline{BC} and contains the center of the inscribed circle of triangle ABC.ABC. Then DE=mn,DE = \frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

知识点：相似内切圆、内心与内切圆半径
难度评级：2390
小提示：

三角形 ADEADE 与 ABCABC 相似，相似比为 h−rh\frac{h - r}{h}，其中 hh 是从 AA 作的高，rr 是内切圆半径。

Triangles ADEADE and ABCABC are similar with ratio h−rh,\frac{h - r}{h}, where hh is the height from AA and rr is the inradius

大提示：

r=Ksr = \frac{K}{s}，且 h=2KBCh = \frac{2K}{BC}，其中 KK 是面积，所以 rh=BC2s\frac{r}{h} = \frac{BC}{2s}，不需要真的算出 KK。

r=Ksr = \frac{K}{s} and h=2KBCh = \frac{2K}{BC} where KK is the area, so rh=BC2s\frac{r}{h} = \frac{BC}{2s} — no need to compute KK

解答：

因为 DE‾∥BC‾\overline{DE} \parallel \overline{BC}，三角形 ADEADE 与 ABCABC 相似，相似比等于从 AA 作出的两条高之比。直线 DEDE 经过内心，而内心到 BCBC 的距离是内切圆半径 rr，所以相似比为 h−rh=1−rh\frac{h - r}{h} = 1 - \frac{r}{h}，其中 hh 是从 AA 到 BC‾\overline{BC} 的高。

若 KK 为面积，s=21+22+202=632s = \frac{21 + 22 + 20}{2} = \frac{63}{2} 为半周长，则 r=Ksr = \frac{K}{s}，且 h=2K20h = \frac{2K}{20}，因此 rh=202s=2063。\frac{r}{h} = \frac{20}{2s} = \frac{20}{63}\text{。}

因此 DE=20(1−2063)DE = 20\left(1 - \frac{20}{63}\right) =20⋅4363= 20 \cdot \frac{43}{63} =86063= \frac{860}{63}，该分数已最简，所以 m+n=860+63=923m + n = 860 + 63 = 923。

Since DE‾∥BC‾,\overline{DE} \parallel \overline{BC}, triangles ADEADE and ABCABC are similar, and the ratio equals the ratio of their heights from A.A. The line DEDE passes through the incenter, which sits at height rr (the inradius) above BC,BC, so the ratio is h−rh=1−rh,\frac{h - r}{h} = 1 - \frac{r}{h}, where hh is the height from AA to BC‾.\overline{BC}.

If KK is the area and s=21+22+202=632s = \frac{21 + 22 + 20}{2} = \frac{63}{2} the semiperimeter, then r=Ksr = \frac{K}{s} and h=2K20,h = \frac{2K}{20}, so rh=202s=2063.\frac{r}{h} = \frac{20}{2s} = \frac{20}{63}.

Therefore DE=20(1−2063)DE = 20\left(1 - \frac{20}{63}\right) =20⋅4363= 20 \cdot \frac{43}{63} =86063,= \frac{860}{63}, which is in lowest terms, and m+n=860+63=923.m + n = 860 + 63 = 923.

8.

如果正整数 NN 的 77 进制表示中的数字，按 1010 进制数来读，等于 NN 的两倍，则称 NN 为一个 77–1010 双数。例如，5151 是一个 77–1010 双数，因为它的 77 进制表示为 102102。最大的 77–1010 双数是多少？

Call a positive integer NN a 77–1010 double if the digits of the base-77 representation of NN form a base-1010 number that is twice N.N. For example, 5151 is a 77–1010 double because its base-77 representation is 102.102. What is the largest 77–1010 double?

知识点：进制数字极限情形界定
难度评级：2430
小提示：

若 NN 的 77 进制数字为 d2d1d0d_2 d_1 d_0，条件为 100d2+10d1+d0100 d_2 + 10 d_1 + d_0 =2(49d2+7d1+d0)= 2(49 d_2 + 7 d_1 + d_0)。

If NN has base-77 digits d2d1d0,d_2 d_1 d_0, the condition says 100d2+10d1+d0100 d_2 + 10 d_1 + d_0 =2(49d2+7d1+d0)= 2(49 d_2 + 7 d_1 + d_0)

大提示：

这会化简为 2d2=4d1+d02 d_2 = 4 d_1 + d_0。四位数不可能，因为 103−2⋅73=31410^3 - 2 \cdot 7^3 = 314，大于其余各项所能抵消的总量。

That simplifies to 2d2=4d1+d0.2 d_2 = 4 d_1 + d_0. A fourth digit is impossible because 103−2⋅73=31410^3 - 2 \cdot 7^3 = 314 is bigger than the other terms could balance.

解答：

设 NN 的 77 进制数字为 dk…d1d0d_k \ldots d_1 d_0。条件是 ∑di 10i=2∑di 7i\sum d_i \, 10^i = 2 \sum d_i \, 7^i，也就是 ∑di (10i−2⋅7i)=0\sum d_i \,(10^i - 2 \cdot 7^i) = 0。系数 10i−2⋅7i10^i - 2 \cdot 7^i 在 i=0,1,2,3i = 0, 1, 2, 3 时分别为 −1-1、−4-4、22、314314。每个满足 i≥3i\ge3 的系数都是正数且至少为 314314。因此，若某个 i≥3i\ge3 位上的数字非零，它的正贡献就无法被两个负项抵消；这两个负项的绝对值之和至多为 4⋅6+6=304\cdot6+6=30。所以 NN 的 77 进制表示至多有三位。

对三位数，条件变为 2d2=4d1+d02 d_2 = 4 d_1 + d_0。为了最大化 N=49d2+7d1+d0N = 49 d_2 + 7 d_1 + d_0，取 d2=6d_2 = 6，于是 4d1+d0=124 d_1 + d_0 = 12；使 7d1+d07 d_1 + d_0 最大的是 d1=3d_1 = 3，d0=0d_0 = 0。

因此 N=49⋅6+7⋅3=315N = 49 \cdot 6 + 7 \cdot 3 = 315，它的 77 进制表示为 630=2⋅315630 = 2 \cdot 315。

Suppose NN has base-77 digits dk…d1d0.d_k \ldots d_1 d_0. The condition is ∑di 10i=2∑di 7i,\sum d_i \, 10^i = 2 \sum d_i \, 7^i, that is ∑di (10i−2⋅7i)=0.\sum d_i \,(10^i - 2 \cdot 7^i) = 0. The coefficients 10i−2⋅7i10^i - 2 \cdot 7^i for i=0,1,2,3i = 0, 1, 2, 3 are −1,-1, −4,-4, 2,2, 314.314. Every coefficient for i≥3i\ge3 is positive and at least 314.314. Thus if any digit in a place i≥3i\ge3 were nonzero, its positive contribution could not be canceled by the two negative terms, whose total magnitude is at most 4⋅6+6=30.4\cdot6+6=30. So NN has at most three base-77 digits.

For three digits the condition reads 2d2=4d1+d0.2 d_2 = 4 d_1 + d_0. To maximize N=49d2+7d1+d0,N = 49 d_2 + 7 d_1 + d_0, take d2=6,d_2 = 6, so 4d1+d0=12;4 d_1 + d_0 = 12; the largest value of 7d1+d07 d_1 + d_0 comes from d1=3,d_1 = 3, d0=0.d_0 = 0.

Thus N=49⋅6+7⋅3=315,N = 49 \cdot 6 + 7 \cdot 3 = 315, whose base-77 representation is 630=2⋅315.630 = 2 \cdot 315.

9.

在三角形 ABCABC 中，AB=13AB = 13、BC=15BC = 15、CA=17CA = 17。点 DD 在 AB‾\overline{AB} 上，点 EE 在 BC‾\overline{BC} 上，点 FF 在 CA‾\overline{CA} 上。令 AD=p⋅ABAD = p \cdot AB、BE=q⋅BCBE = q \cdot BC、CF=r⋅CACF = r \cdot CA，其中 pp、qq、rr 为正数，且满足 p+q+r=23p + q + r = \frac{2}{3} 和 p2+q2+r2=25p^2 + q^2 + r^2 = \frac{2}{5}。三角形 DEFDEF 与三角形 ABCABC 的面积比可写成 mn\frac{m}{n}，其中 mm 和 nn 是互质正整数。求 m+nm + n。

In triangle ABC,ABC, AB=13,AB = 13, BC=15,BC = 15, and CA=17.CA = 17. Point DD is on AB‾,\overline{AB}, EE is on BC‾,\overline{BC}, and FF is on CA‾.\overline{CA}. Let AD=p⋅AB,AD = p \cdot AB, BE=q⋅BC,BE = q \cdot BC, and CF=r⋅CA,CF = r \cdot CA, where p,p, q,q, and rr are positive and satisfy p+q+r=23p + q + r = \frac{2}{3} and p2+q2+r2=25.p^2 + q^2 + r^2 = \frac{2}{5}. The ratio of the area of triangle DEFDEF to the area of triangle ABCABC can be written in the form mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

知识点：面积比对称性（代数）代数变形
难度评级：2560
小提示：

减去三个角上的三角形：[DEF][ABC]=1\frac{[DEF]}{[ABC]} = 1 −p(1−r)- p(1-r) −q(1−p)- q(1-p) −r(1−q)- r(1-q)。

Subtract the three corner triangles: [DEF][ABC]=1\frac{[DEF]}{[ABC]} = 1 −p(1−r)- p(1-r) −q(1−p)- q(1-p) −r(1−q)- r(1-q)

大提示：

pq+qr+rp=(p+q+r)2−(p2+q2+r2)2pq + qr + rp = \frac{(p+q+r)^2 - (p^2+q^2+r^2)}{2}。

pq+qr+rp=(p+q+r)2−(p2+q2+r2)2pq + qr + rp = \frac{(p+q+r)^2 - (p^2+q^2+r^2)}{2}

解答：

每个角上的小三角形面积都是对应边长比例的乘积：[ADF]=p(1−r)[ABC][ADF] = p(1-r)[ABC]、[BED]=q(1−p)[ABC][BED] = q(1-p)[ABC]、[CFE]=r(1−q)[ABC][CFE] = r(1-q)[ABC]。这里使用的是共用角下的 12xysin⁡θ\frac{1}{2}xy\sin\theta 面积公式。相减得 [DEF][ABC]=1−p(1−r)−q(1−p)−r(1−q)=1−(p+q+r)+(pq+qr+rp)。 \begin{aligned} \frac{[DEF]}{[ABC]} &= 1 - p(1-r) \\ &\quad {}- q(1-p) - r(1-q) \\ &= 1 - (p+q+r) \\ &\quad {}+ (pq+qr+rp) \end{aligned}\text{。}

由题给条件，pq+qr+rp=(23)2−252pq + qr + rp = \frac{(\frac{2}{3})^2 - \frac{2}{5}}{2} =49−252= \frac{\frac{4}{9} - \frac{2}{5}}{2} =145= \frac{1}{45}。

因此面积比为 1−23+145=16451 - \frac{2}{3} + \frac{1}{45} = \frac{16}{45}，所以 m+n=16+45=61m + n = 16 + 45 = 61。

Each corner triangle’s area is a product of side fractions: [ADF]=p(1−r)[ABC],[ADF] = p(1-r)[ABC], [BED]=q(1−p)[ABC],[BED] = q(1-p)[ABC], and [CFE]=r(1−q)[ABC],[CFE] = r(1-q)[ABC], using the formula 12xysin⁡θ\frac{1}{2}xy\sin\theta on the shared angles. Subtracting, [DEF][ABC]=1−p(1−r)−q(1−p)−r(1−q)=1−(p+q+r)+(pq+qr+rp). \begin{aligned} \frac{[DEF]}{[ABC]} &= 1 - p(1-r) \\ &\quad {}- q(1-p) - r(1-q) \\ &= 1 - (p+q+r) \\ &\quad {}+ (pq+qr+rp). \end{aligned}

From the given values, pq+qr+rp=(23)2−252pq + qr + rp = \frac{(\frac{2}{3})^2 - \frac{2}{5}}{2} =49−252= \frac{\frac{4}{9} - \frac{2}{5}}{2} =145.= \frac{1}{45}.

Therefore the ratio is 1−23+145=1645,1 - \frac{2}{3} + \frac{1}{45} = \frac{16}{45}, and m+n=16+45=61.m + n = 16 + 45 = 61.

10.

令 S\mathcal{S} 为所有坐标 xx、yy、zz 均为整数，且满足 0≤x≤20 \le x \le 2、0≤y≤30 \le y \le 3、0≤z≤40 \le z \le 4 的点的集合。从 S\mathcal{S} 中随机选取两个不同的点。它们确定的线段中点也属于 S\mathcal{S} 的概率为 mn\frac{m}{n}，其中 mm 和 nn 是互质正整数。求 m+nm + n。

Let S\mathcal{S} be the set of points whose coordinates x,x, y,y, and zz are integers that satisfy 0≤x≤2,0 \le x \le 2, 0≤y≤3,0 \le y \le 3, and 0≤z≤4.0 \le z \le 4. Two distinct points are randomly chosen from S.\mathcal{S}. The probability that the midpoint of the segment they determine also belongs to S\mathcal{S} is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

知识点：奇偶性基本概率数对计数格点
难度评级：2500
小提示：

中点坐标为整数，当且仅当两个点在每个坐标上的奇偶性都相同。

The midpoint has integer coordinates exactly when the two points have the same parity in each coordinate

大提示：

逐个坐标计算同奇偶性的有序对数（xx 有 55 种、yy 有 88 种、zz 有 1313 种），再减去两点重合的点对。

Count ordered same-parity pairs coordinate by coordinate (55 ways for x,x, 88 for y,y, 1313 for zz), then subtract the pairs where both points coincide

解答：

中点是格点，当且仅当所选两点在每个坐标上的奇偶性相同。先允许两点相同，按坐标计算有序点对。对 x∈{0,1,2}x \in \{0, 1, 2\}，有 22 个偶数值和 11 个奇数值，因此同奇偶有序对数为 22+12=52^2 + 1^2 = 5。对 y∈{0,…,3}y \in \{0, \ldots, 3\}，得到 22+22=82^2 + 2^2 = 8。对 z∈{0,…,4}z \in \{0, \ldots, 4\}，得到 32+22=133^2 + 2^2 = 13。

因此共有 5⋅8⋅13=5205 \cdot 8 \cdot 13 = 520 个有序点对，其中包括 6060 个两点相同的点对，所以有 520−60=460520 - 60 = 460 个不同点的有序点对，即 230230 个无序点对。全部无序点对数为 (602)=1770\binom{60}{2} = 1770。

概率为 2301770=23177\frac{230}{1770} = \frac{23}{177}。由于 177=3⋅59177 = 3 \cdot 59，该分数已最简。因此 m+n=23+177=200m + n = 23 + 177 = 200。

The midpoint is a lattice point exactly when the two chosen points agree in parity in each coordinate. Count ordered pairs (allowing equality) coordinate by coordinate. For x∈{0,1,2}x \in \{0, 1, 2\} there are 22 even and 11 odd values, giving 22+12=52^2 + 1^2 = 5 same-parity ordered pairs. For y∈{0,…,3}:y \in \{0, \ldots, 3\}: 22+22=8.2^2 + 2^2 = 8. For z∈{0,…,4}:z \in \{0, \ldots, 4\}: 32+22=13.3^2 + 2^2 = 13.

That gives 5⋅8⋅13=5205 \cdot 8 \cdot 13 = 520 ordered pairs, including the 6060 pairs where the two points are equal, so 520−60=460520 - 60 = 460 ordered pairs of distinct points, or 230230 unordered pairs. The total number of unordered pairs is (602)=1770.\binom{60}{2} = 1770.

The probability is 2301770=23177,\frac{230}{1770} = \frac{23}{177}, and since 177=3⋅59,177 = 3 \cdot 59, this is in lowest terms. Thus m+n=23+177=200.m + n = 23 + 177 = 200.

11.

在一个有 55 行 NN 列的长方形点阵中，点从最上面一行开始，按从左到右的顺序逐行连续编号。因此最上面一行编号为 11 到 NN，第二行编号为 N+1N + 1 到 2N2N，依此类推。选取五个点 P1P_1、P2P_2、P3P_3、P4P_4、P5P_5，使得每个 PiP_i 位于第 ii 行。令 xix_i 为点 PiP_i 对应的编号。现在改为从第一列开始，按从上到下的顺序逐列连续编号。令 yiy_i 为重新编号后点 PiP_i 对应的编号。

已知 x1=y2x_1 = y_2、x2=y1x_2 = y_1、x3=y4x_3 = y_4、x4=y5x_4 = y_5、x5=y3x_5 = y_3。求 NN 的最小可能值。

In a rectangular array of points, with 55 rows and NN columns, the points are numbered consecutively from left to right beginning with the top row. Thus the top row is numbered 11 through N,N, the second row is numbered N+1N + 1 through 2N,2N, and so forth. Five points, P1,P_1, P2,P_2, P3,P_3, P4,P_4, and P5,P_5, are selected so that each PiP_i is in row i.i. Let xix_i be the number associated with Pi.P_i. Now renumber the array consecutively from top to bottom, beginning with the first column. Let yiy_i be the number associated with PiP_i after renumbering.

It is found that x1=y2,x_1 = y_2, x2=y1,x_2 = y_1, x3=y4,x_3 = y_4, x4=y5,x_4 = y_5, and x5=y3.x_5 = y_3. Find the smallest possible value of N.N.

知识点：方程组模运算
难度评级：2990
小提示：

若 PiP_i 在第 cic_i 列，则 xi=(i−1)N+cix_i = (i-1)N + c_i，且 yi=5(ci−1)+iy_i = 5(c_i - 1) + i。

If PiP_i is in column ci,c_i, then xi=(i−1)N+cix_i = (i-1)N + c_i and yi=5(ci−1)+iy_i = 5(c_i - 1) + i

大提示：

前两个方程给出 N=24c2−19N = 24c_2 - 19；另外三个方程会给出关于 c2c_2 的整除条件，再模 3131 求解。

The first two equations give N=24c2−19;N = 24c_2 - 19; the other three force a divisibility condition on c2.c_2. Solve it modulo 31.31.

解答：

设 PiP_i 位于第 cic_i 列。则 xi=(i−1)N+cix_i = (i-1)N + c_i，且 yi=5(ci−1)+iy_i = 5(c_i - 1) + i。五个条件变为 c1=5c2−3,N+c2=5c1−4,2N+c3=5c4−1， \begin{aligned} c_1 &= 5c_2 - 3, \\ N + c_2 &= 5c_1 - 4, \\ 2N + c_3 &= 5c_4 - 1 \end{aligned}\text{，}3N+c4=5c5,4N+c5=5c3−2。 \begin{aligned} 3N + c_4 &= 5c_5, \\ 4N + c_5 &= 5c_3 - 2 \end{aligned}\text{。}

将 c1=5c2−3c_1 = 5c_2 - 3 代入第二个方程，得 N=24c2−19N = 24c_2 - 19。从后三个方程消去 c3c_3 和 c4c_4，得到 124c5=89N+7124 c_5 = 89N + 7。再代入 N=24c2−19N = 24c_2 - 19 并化简，即 534c2−421534 c_2 - 421 能被 3131 整除，也就是 7c2≡18(mod31)7 c_2 \equiv 18 \pmod{31}。其最小正整数解为 c2=7c_2 = 7。

于是 N=24⋅7−19=149N = 24 \cdot 7 - 19 = 149。回代可得有效列号 (c1,…,c5)=(c_1, \ldots, c_5)= (32,7,141,88,107)(32, 7, 141, 88, 107)，都不超过 149149。对应的编号对为 (x1,y2)=(32,32)(x_1,y_2)=(32,32)、(x2,y1)=(156,156)(x_2,y_1)=(156,156)、(x3,y4)=(439,439)(x_3,y_4)=(439,439)、(x4,y5)=(535,535)(x_4,y_5)=(535,535)、(x5,y3)=(703,703)(x_5,y_3)=(703,703)。所以 NN 的最小可能值为 149149。

Let PiP_i sit in column ci.c_i. Then xi=(i−1)N+cix_i = (i-1)N + c_i and yi=5(ci−1)+i.y_i = 5(c_i - 1) + i. The five conditions become c1=5c2−3,N+c2=5c1−4,2N+c3=5c4−1, \begin{aligned} c_1 &= 5c_2 - 3, \\ N + c_2 &= 5c_1 - 4, \\ 2N + c_3 &= 5c_4 - 1, \end{aligned} 3N+c4=5c5,4N+c5=5c3−2. \begin{aligned} 3N + c_4 &= 5c_5, \\ 4N + c_5 &= 5c_3 - 2. \end{aligned}

Substituting c1=5c2−3c_1 = 5c_2 - 3 into the second equation gives N=24c2−19.N = 24c_2 - 19. Eliminating c3c_3 and c4c_4 from the last three equations yields 124c5=89N+7.124 c_5 = 89N + 7. Substituting N=24c2−19N = 24c_2 - 19 and reducing, 534c2−421534 c_2 - 421 is divisible by 31,31, i.e. 7c2≡18(mod31),7 c_2 \equiv 18 \pmod{31}, whose smallest positive solution is c2=7.c_2 = 7.

Then N=24⋅7−19=149,N = 24 \cdot 7 - 19 = 149, and back-substituting gives valid columns (c1,…,c5)=(c_1, \ldots, c_5)= (32,7,141,88,107),(32, 7, 141, 88, 107), all at most 149.149. Indeed the corresponding pairs are (x1,y2)=(32,32),(x_1,y_2)=(32,32), (x2,y1)=(156,156),(x_2,y_1)=(156,156), (x3,y4)=(439,439),(x_3,y_4)=(439,439), (x4,y5)=(535,535),(x_4,y_5)=(535,535), and (x5,y3)=(703,703).(x_5,y_3)=(703,703). So the smallest possible NN is 149.149.

12.

一个球内切于顶点为 A=(6,0,0)A = (6, 0, 0)、B=(0,4,0)B = (0, 4, 0)、C=(0,0,2)C = (0, 0, 2) 和 D=(0,0,0)D = (0, 0, 0) 的四面体。该球的半径为 mn\frac{m}{n}，其中 mm 和 nn 是互质的正整数。求 m+nm + n。

A sphere is inscribed in the tetrahedron whose vertices are A=(6,0,0),A = (6, 0, 0), B=(0,4,0),B = (0, 4, 0), C=(0,0,2),C = (0, 0, 2), and D=(0,0,0).D = (0, 0, 0). The radius of the sphere is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

知识点：立体几何内切圆、内心与内切圆半径体积向量
难度评级：2560
小提示：

把四面体分解为以内切球球心为顶点的四个棱锥：V=13rSV = \frac{1}{3} r S，其中 SS 是总表面积。

Decompose the tetrahedron into four pyramids with apex at the insphere center: V=13rSV = \frac{1}{3} r S where SS is the total surface area

大提示：

V=16⋅6⋅4⋅2V = \frac{1}{6} \cdot 6 \cdot 4 \cdot 2，三个直角面的面积为 1212、66、44，面 ABCABC 的面积为 1414。

V=16⋅6⋅4⋅2,V = \frac{1}{6} \cdot 6 \cdot 4 \cdot 2, the three right-angle faces have areas 12,12, 6,6, 4,4, and face ABCABC has area 1414

解答：

连接内切球球心与四个顶点，会把四面体分成四个以各面为底、高为 rr 的棱锥，所以 V=13rSV = \frac{1}{3} r S，即 r=3VSr = \frac{3V}{S}，其中 SS 是总表面积。

这里 V=16⋅6⋅4⋅2=8V = \frac{1}{6} \cdot 6 \cdot 4 \cdot 2 = 8。三个在坐标平面上的面是直角三角形，面积分别为 12⋅6⋅4=12\frac{1}{2} \cdot 6 \cdot 4 = 12、12⋅6⋅2=6\frac{1}{2} \cdot 6 \cdot 2 = 6、12⋅4⋅2=4\frac{1}{2} \cdot 4 \cdot 2 = 4。对面 ABCABC，AB→=(−6,4,0)\overrightarrow{AB} = (-6, 4, 0) 与 AC→=(−6,0,2)\overrightarrow{AC} = (-6, 0, 2) 的叉积为 (8,12,24)(8, 12, 24)，长度为 422+32+62=284\sqrt{2^2 + 3^2 + 6^2} = 28，所以该面的面积为 1414。

因此 S=12+6+4+14=36S = 12 + 6 + 4 + 14 = 36，且 r=2436=23r = \frac{24}{36} = \frac{2}{3}，所以 m+n=2+3=5m + n = 2 + 3 = 5。

Connecting the incenter to the four vertices splits the tetrahedron into four pyramids of height rr over the faces, so V=13rS,V = \frac{1}{3} r S, i.e. r=3VS,r = \frac{3V}{S}, where SS is the total surface area.

Here V=16⋅6⋅4⋅2=8.V = \frac{1}{6} \cdot 6 \cdot 4 \cdot 2 = 8. The three faces on the coordinate planes are right triangles with areas 12⋅6⋅4=12,\frac{1}{2} \cdot 6 \cdot 4 = 12, 12⋅6⋅2=6,\frac{1}{2} \cdot 6 \cdot 2 = 6, and 12⋅4⋅2=4.\frac{1}{2} \cdot 4 \cdot 2 = 4. For face ABC,ABC, the cross product of AB→=(−6,4,0)\overrightarrow{AB} = (-6, 4, 0) and AC→=(−6,0,2)\overrightarrow{AC} = (-6, 0, 2) is (8,12,24),(8, 12, 24), with length 422+32+62=28,4\sqrt{2^2 + 3^2 + 6^2} = 28, so that face has area 14.14.

Then S=12+6+4+14=36S = 12 + 6 + 4 + 14 = 36 and r=2436=23,r = \frac{24}{36} = \frac{2}{3}, giving m+n=2+3=5.m + n = 2 + 3 = 5.

13.

在某个圆中，对应 dd 度弧的弦长为 2222 厘米，且对应 2d2d 度弧的弦比对应 3d3d 度弧的弦长 2020 厘米，其中 d<120d \lt 120。对应 3d3d 度弧的弦长为 −m+n-m + \sqrt{n} 厘米，其中 mm 和 nn 是正整数。求 m+nm + n。

In a certain circle, the chord of a dd-degree arc is 2222 centimeters long, and the chord of a 2d2d-degree arc is 2020 centimeters longer than the chord of a 3d3d-degree arc, where d<120.d \lt 120. The length of the chord of a 3d3d-degree arc is −m+n-m + \sqrt{n} centimeters, where mm and nn are positive integers. Find m+n.m + n.

知识点：弦三角恒等式二次方程
难度评级：2990
小提示：

对应 θ\theta 度弧的弦长为 2Rsin⁡θ22R\sin\frac{\theta}{2}。设 t=d2t = \frac{d}{2}，并使用 sin⁡2t=2sin⁡tcos⁡t\sin 2t = 2\sin t\cos t、sin⁡3t=sin⁡t (4cos⁡2t−1)\sin 3t = \sin t\,(4\cos^2 t - 1)。

A chord of a θ\theta-degree arc has length 2Rsin⁡θ2.2R\sin\frac{\theta}{2}. Set t=d2t = \frac{d}{2} and use sin⁡2t=2sin⁡tcos⁡t,\sin 2t = 2\sin t\cos t, sin⁡3t=sin⁡t (4cos⁡2t−1).\sin 3t = \sin t\,(4\cos^2 t - 1).

大提示：

全部除以 2Rsin⁡t=222R \sin t = 22，得到 44cos⁡2t−22cos⁡t−1=044\cos^2 t - 22\cos t - 1 = 0，并注意 3d3d 度弧的弦长等于 44cos⁡t−2044 \cos t - 20。

Divide everything by 2Rsin⁡t=222R \sin t = 22 to get 44cos⁡2t−22cos⁡t−1=0,44\cos^2 t - 22\cos t - 1 = 0, and note the 3d3d-degree chord equals 44cos⁡t−2044 \cos t - 20

解答：

在半径为 RR 的圆中，对应 θ\theta 度弧的弦长为 2Rsin⁡θ22R\sin\frac{\theta}{2}。令 t=d2t = \frac{d}{2}，则三条弦长分别为 2Rsin⁡t=222R\sin t = 22、2Rsin⁡2t2R \sin 2t 和 2Rsin⁡3t2R \sin 3t。利用 sin⁡2t=2sin⁡tcos⁡t\sin 2t = 2 \sin t \cos t 以及 sin⁡3t=sin⁡t (4cos⁡2t−1)\sin 3t = \sin t\,(4\cos^2 t - 1)，对应 2d2d 度和 3d3d 度弧的弦长分别为 22⋅2cos⁡t=44cos⁡t22 \cdot 2\cos t = 44 \cos t 与 22 (4cos⁡2t−1)22\,(4\cos^2 t - 1)。

“2d2d 弦比 3d3d 弦长 2020”这一条件变为 44cos⁡t=22 (4cos⁡2t−1)+2044\cos t = 22\,(4\cos^2 t - 1) + 20，化简得 44cos⁡2t−22cos⁡t−1=044 \cos^2 t - 22 \cos t - 1 = 0。由此 4cos⁡2t=2cos⁡t+1114\cos^2 t = 2\cos t + \frac{1}{11}，所以 3d3d 弦长为 22(2cos⁡t+111−1)22\left(2\cos t + \frac{1}{11} - 1\right) =44cos⁡t−20= 44 \cos t - 20。

解二次方程，cos⁡t=22+484+17688=11+16544\cos t = \frac{22 + \sqrt{484 + 176}}{88} = \frac{11 + \sqrt{165}}{44}（取 ++ 根，因为 d<120d \lt 120 意味着 t<60∘t \lt 60^\circ，所以 cos⁡t>12\cos t \gt \frac{1}{2}）。于是 3d3d 弦长为 44cos⁡t−2044\cos t - 20 =11+165−20= 11 + \sqrt{165} - 20 =−9+165= -9 + \sqrt{165}，所以 m+n=9+165=174m + n = 9 + 165 = 174。

A chord subtending a θ\theta-degree arc in a circle of radius RR has length 2Rsin⁡θ2.2R\sin\frac{\theta}{2}. Write t=d2,t = \frac{d}{2}, so the three chords are 2Rsin⁡t=22,2R\sin t = 22, 2Rsin⁡2t,2R \sin 2t, and 2Rsin⁡3t.2R \sin 3t. Using sin⁡2t=2sin⁡tcos⁡t\sin 2t = 2 \sin t \cos t and sin⁡3t=sin⁡t (4cos⁡2t−1),\sin 3t = \sin t\,(4\cos^2 t - 1), the chords of the 2d2d- and 3d3d-degree arcs are 22⋅2cos⁡t=44cos⁡t22 \cdot 2\cos t = 44 \cos t and 22 (4cos⁡2t−1).22\,(4\cos^2 t - 1).

The condition “the 2d2d-chord is 2020 longer than the 3d3d-chord” becomes 44cos⁡t=22 (4cos⁡2t−1)+20,44\cos t = 22\,(4\cos^2 t - 1) + 20, which simplifies to 44cos⁡2t−22cos⁡t−1=0.44 \cos^2 t - 22 \cos t - 1 = 0. From this, 4cos⁡2t=2cos⁡t+111,4\cos^2 t = 2\cos t + \frac{1}{11}, so the 3d3d-chord equals 22(2cos⁡t+111−1)22\left(2\cos t + \frac{1}{11} - 1\right) =44cos⁡t−20.= 44 \cos t - 20.

Solving the quadratic, cos⁡t=22+484+17688=11+16544\cos t = \frac{22 + \sqrt{484 + 176}}{88} = \frac{11 + \sqrt{165}}{44} (the ++ root since d<120d \lt 120 means t<60∘,t \lt 60^\circ, so cos⁡t>12\cos t \gt \frac{1}{2}). Then the 3d3d-chord is 44cos⁡t−2044\cos t - 20 =11+165−20= 11 + \sqrt{165} - 20 =−9+165,= -9 + \sqrt{165}, giving m+n=9+165=174.m + n = 9 + 165 = 174.

14.

一名邮递员给榆树街东侧的十九户人家送信。邮递员注意到，同一天没有相邻的两户同时收到信，但同一天也从不会有连续超过两户没有收到信。共有多少种不同的送信模式？

A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible?

知识点：递推计数有限制的排列
难度评级：2760
小提示：

用二元串表示是否送信，要求没有 1111，也没有 000000。按有效串以 11、1010 或 100100 结尾分类。

Encode delivery as a binary string with no 1111 and no 000.000. Classify valid strings by whether they end in 1,1, 10,10, or 100.100.

大提示：

令 An,Bn,CnA_n, B_n, C_n 分别计数这些结尾，则 An=Bn−1+Cn−1A_n = B_{n-1} + C_{n-1}、Bn=An−1B_n = A_{n-1}、Cn=Bn−1C_n = B_{n-1}。迭代到 n=19n = 19。

With An,Bn,CnA_n, B_n, C_n counting those endings: An=Bn−1+Cn−1,A_n = B_{n-1} + C_{n-1}, Bn=An−1,B_n = A_{n-1}, Cn=Bn−1.C_n = B_{n-1}. Iterate up to n=19.n = 19.

解答：

用 11 表示某户收到信，00 表示没有收到信。有效模式是长度为 1919 的二元串，其中没有连续两个 11，也没有连续三个 00。令 AnA_n、BnB_n、CnC_n 分别表示长度为 nn 的有效串中，以 11 结尾、恰好以一个 00 结尾、恰好以两个 00 结尾的数量。一个 11 可以接在任一种以 00 结尾的串后，一个单独的 00 可以接在 11 后，第二个 00 可以接在一个 00 后：An=Bn−1+Cn−1,Bn=An−1,Cn=Bn−1。 \begin{aligned} A_n &= B_{n-1} + C_{n-1}, \\ B_n &= A_{n-1}, \\ C_n &= B_{n-1} \end{aligned}\text{。}

从 A1=B1=1A_1 = B_1 = 1、C1=0C_1 = 0 开始迭代，总数 An+Bn+CnA_n + B_n + C_n 依次为 22、33、44、55、77、99、1212、1616、2121、2828、3737、4949、6565、8686、114114、151151、200200、265265、351351。

当 n=19n = 19 时，数量为 351351。

Write 11 for a house that gets mail and 00 for one that does not. Valid patterns are binary strings of length 1919 with no two consecutive 11s and no three consecutive 00s. Let An,A_n, Bn,B_n, CnC_n count valid length-nn strings ending in 1,1, in exactly one 0,0, and in exactly two 00s. A 11 may follow either kind of 00-ending, a single 00 may follow a 1,1, and a second 00 may follow a single 0:0: An=Bn−1+Cn−1,Bn=An−1,Cn=Bn−1. \begin{aligned} A_n &= B_{n-1} + C_{n-1}, \\ B_n &= A_{n-1}, \\ C_n &= B_{n-1}. \end{aligned}

Starting from A1=B1=1,A_1 = B_1 = 1, C1=0C_1 = 0 and iterating, the totals An+Bn+CnA_n + B_n + C_n run 2,2, 3,3, 4,4, 5,5, 7,7, 9,9, 12,12, 16,16, 21,21, 28,28, 37,37, 49,49, 65,65, 86,86, 114,114, 151,151, 200,200, 265,265, 351.351.

For n=19n = 19 the count is 351.351.

15.

将数字 11，22，33，44，55，66，77 和 88 随机写在正八面体的各个面上，每个面写不同的数字。若没有两个连续的数字（其中 88 和 11 也视为连续）写在共边的面上，其概率为 mn\frac{m}{n}，其中 mm 和 nn 是互质正整数。求 m+nm + n。

The numbers 1,1, 2,2, 3,3, 4,4, 5,5, 6,6, 7,7, and 88 are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where 88 and 11 are considered to be consecutive, are written on faces that share an edge is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

知识点：多面体图论基本概率分类讨论
难度评级：3270
小提示：

八面体的面对应立方体的顶点，两个面共边当且仅当对应的立方体顶点相邻。循环 1,2,…,8,11, 2, \ldots, 8, 1 必须只使用立方体的对角线。

The faces of the octahedron correspond to the vertices of a cube, with faces sharing an edge exactly when the cube vertices are adjacent. The cycle 1,2,…,8,11, 2, \ldots, 8, 1 must use only cube diagonals.

大提示：

这 1616 条对角线组成两个四面体加上 44 条长对角线，且每个顶点恰在一条长对角线上，所以有效的 88 环使用 22 条或 44 条长对角线。

The 1616 diagonals form two tetrahedra plus 44 long diagonals, and each vertex lies on exactly one long diagonal, so a valid 88-cycle uses either 22 or 44 long diagonals

解答：

转到对偶立方体：八面体的面对应立方体的顶点，两个面共边当且仅当对应的立方体顶点相邻。按照 1,2,…,81, 2, \ldots, 8 再回到 11 的顺序，会沿所有立方体顶点走出一个闭合的 88 步回路，而条件要求每一步都是对角线（两个内接四面体的边之一，或 44 条长空间对角线之一）。这样的对角线共有 1616 条。

每个顶点恰在一条长对角线上，所以回路不能连续走两条长对角线；并且只有通过长对角线才能在两个四面体之间切换。因此回路要么使用 44 条长对角线并与四面体边交替，要么使用 22 条长对角线，在每个四面体中由 33 条边的路径连接。第一种情况中，在每个四面体中选择一对对边（3⋅23 \cdot 2 种）得到 66 个八边形，每个可按 8⋅28 \cdot 2 种方式追踪为排列，共有 9696 种。第二种情况中，在一个四面体中选一条 33 边路径有 4!=244! = 24 种，另一个四面体中的返回路径随后只剩 22 种选择，共 8⋅24⋅2=3848 \cdot 24 \cdot 2 = 384 种排列。

因此在 8!=403208! = 40320 种标号中，有 96+384=48096 + 384 = 480 种满足条件，概率为 48040320=184\frac{480}{40320} = \frac{1}{84}。所以 m+n=1+84=85m + n = 1 + 84 = 85。

Pass to the dual cube: the octahedron’s faces correspond to a cube’s vertices, and two faces share an edge exactly when the corresponding cube vertices are adjacent. Following the numbers 1,2,…,81, 2, \ldots, 8 and back to 11 traces a closed 88-step circuit through all the cube’s vertices, and the requirement is that every step is a diagonal (an edge of one of the two inscribed tetrahedra, or one of the 44 long space diagonals). There are 1616 such diagonals.

Each vertex lies on exactly one long diagonal, so the circuit cannot take two long diagonals in a row, and switching between the two tetrahedra is possible only via a long diagonal. Hence the circuit uses either 44 long diagonals alternating with tetrahedron edges, or 22 long diagonals separated by 33-edge paths in each tetrahedron. In the first case, choosing a pair of opposite edges in each tetrahedron (3⋅23 \cdot 2 ways) gives 66 octagons, each traceable as 8⋅28 \cdot 2 permutations: 96.96. In the second case, a 33-edge path in one tetrahedron can be chosen in 4!=244! = 24 ways, and the return path through the other tetrahedron is then forced up to 22 choices, giving 8⋅24⋅2=3848 \cdot 24 \cdot 2 = 384 permutations.

So 96+384=48096 + 384 = 480 of the 8!=403208! = 40320 labelings work, and the probability is 48040320=184.\frac{480}{40320} = \frac{1}{84}. Thus m+n=1+84=85.m + n = 1 + 84 = 85.