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2003 AIME I 真题

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1.

已知 ((3!)!)!3!=k⋅n!\frac{((3!)!)!}{3!} = k \cdot n!，其中 kk 和 nn 是正整数，且 nn 尽可能大。求 k+nk + n。

Given that ((3!)!)!3!=k⋅n!,\frac{((3!)!)!}{3!} = k \cdot n!, where kk and nn are positive integers and nn is as large as possible, find k+n.k + n.

答案：839
知识点：阶乘极限情形界定
难度评级：1670
小提示：

先计算里面的阶乘：3!=63! = 6，所以原式是 720!6\frac{720!}{6}。

Compute the inner factorials: 3!=6,3! = 6, so the expression is 720!6\frac{720!}{6}

大提示：

写成 720!=720⋅719!720! = 720 \cdot 719!，并注意若 n=720n = 720，则 k⋅n!k \cdot n! 会太大。

Write 720!=720⋅719!720! = 720 \cdot 719! and note that n=720n = 720 would make k⋅n!k \cdot n! too large

解答：

因为 3!=63! = 6 且 6!=7206! = 720，原式为 ((3!)!)!3!=720!6=720⋅719!6=120⋅719!。 \begin{aligned} \frac{((3!)!)!}{3!} &= \frac{720!}{6} \\ &= \frac{720 \cdot 719!}{6} \\ &= 120 \cdot 719! \end{aligned}\text{。}

若 nn 为 720720 或更大，则 k⋅n!≥720!k \cdot n! \ge 720!，这超过了 720!6\frac{720!}{6}。因此 nn 的最大可能值是 719719，此时 k=120k = 120，所以 k+n=120+719=839k + n = 120 + 719 = 839。

Since 3!=63! = 6 and 6!=720,6! = 720, the expression is ((3!)!)!3!=720!6=720⋅719!6=120⋅719!. \begin{aligned} \frac{((3!)!)!}{3!} &= \frac{720!}{6} \\ &= \frac{720 \cdot 719!}{6} \\ &= 120 \cdot 719!. \end{aligned}

If nn were 720720 or more, then k⋅n!≥720!,k \cdot n! \ge 720!, which exceeds 720!6.\frac{720!}{6}. So the largest possible value of nn is 719,719, achieved with k=120,k = 120, and k+n=120+719=839.k + n = 120 + 719 = 839.

2.

平面上画出一百个同心圆，半径分别为 11、22、33、…\ldots、100100。半径为 11 的圆内部涂成红色；每两个相邻圆之间围成的区域涂成红色或绿色，且任意两个相邻区域颜色不同。绿色区域的总面积与半径为 100100 的圆面积之比可写为 mn\frac{m}{n}，其中 mm 和 nn 是互质正整数。求 m+nm + n。

One hundred concentric circles with radii 1,1, 2,2, 3,3, …,\ldots, 100100 are drawn in a plane. The interior of the circle of radius 11 is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to the area of the circle of radius 100100 can be expressed as mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

答案：301
知识点：圆环平方差求和
难度评级：1790
小提示：

颜色交替，因此绿色区域是半径 11 与 22 之间、33 与 44 之间的圆环，依此类推。

The colors alternate, so the green regions are the annuli between radii 11 and 2,2, 33 and 4,4, and so on

大提示：

用平方差可得，绿色面积总和为 π (1+2+3+⋯+100)\pi\,(1 + 2 + 3 + \cdots + 100)。

By difference of squares, the green areas total π (1+2+3+⋯+100)\pi\,(1 + 2 + 3 + \cdots + 100)

解答：

从中心向外，区域颜色依次为红、绿、红、绿、…\ldots，所以绿色区域是半径 11 与 22 之间、33 与 44 之间，依此直到 9999 与 100100 之间的圆环。它们的总面积为 π[(22−12)+(42−32)+⋯+(1002−992)]=π[(2+1)+(4+3)+⋯+(100+99)] \begin{aligned} &\scriptsize \pi\left[(2^2 - 1^2) + (4^2 - 3^2) + \cdots + (100^2 - 99^2)\right] \\ &\scriptsize = \pi\left[(2 + 1) + (4 + 3) + \cdots + (100 + 99)\right] \end{aligned} 即 π (1+2+⋯+100)=5050π\pi\,(1 + 2 + \cdots + 100) = 5050\pi。

所求比值为 5050π1002π=101200\frac{5050\pi}{100^2 \pi} = \frac{101}{200}，因此 m+n=101+200=301m + n = 101 + 200 = 301。

The regions alternate red, green, red, green, …\ldots from the center outward, so the green regions are the annuli between radii 11 and 2,2, between 33 and 4,4, and so on up to the annulus between 9999 and 100.100. Their total area is π[(22−12)+(42−32)+⋯+(1002−992)]=π[(2+1)+(4+3)+⋯+(100+99)], \begin{aligned} &\scriptsize \pi\left[(2^2 - 1^2) + (4^2 - 3^2) + \cdots + (100^2 - 99^2)\right] \\ &\scriptsize = \pi\left[(2 + 1) + (4 + 3) + \cdots + (100 + 99)\right], \end{aligned} which is π (1+2+⋯+100)=5050π.\pi\,(1 + 2 + \cdots + 100) = 5050\pi.

The desired ratio is 5050π1002π=101200,\frac{5050\pi}{100^2 \pi} = \frac{101}{200}, so m+n=101+200=301.m + n = 101 + 200 = 301.

3.

设集合 S={8,5,1,13,34,3,21,2}\mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}。苏珊按如下方式列出一张清单：对于 S\mathcal{S} 的每个二元子集，她在清单上写下该子集中较大的元素。求清单上所有数的和。

Let the set S={8,5,1,13,34,3,21,2}.\mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}. Susan makes a list as follows: for each two-element subset of S,\mathcal{S}, she writes on her list the greater of the set’s two elements. Find the sum of the numbers on the list.

答案：484
知识点：数对计数双重计数
难度评级：1840
小提示：

每个元素会因为集合中每个比它小的元素而在表上出现一次。

Each element is written on the list once for every smaller element in the set

大提示：

将集合排成 1,2,3,5,8,13,21,341, 2, 3, 5, 8, 13, 21, 34，并把每个元素乘以比它小的元素个数。

Sort the set as 1,2,3,5,8,13,21,341, 2, 3, 5, 8, 13, 21, 34 and multiply each element by the count of smaller elements

解答：

元素 xx 与集合中每个比它小的元素组成二元子集时，都会成为较大的元素，因此 xx 对总和的贡献次数等于比它小的元素个数。将集合排序为 1,2,3,5,8,13,21,341, 2, 3, 5, 8, 13, 21, 34，表上所有数的和为 0(1)+1(2)+2(3)+3(5)+4(8)+5(13)+6(21)+7(34)=2+6+15+32+65+126+238=484。 \begin{aligned} &0(1) + 1(2) + 2(3) + 3(5) \\ &\quad {}+ 4(8) + 5(13) + 6(21) \\ &\quad {}+ 7(34) \\ &= 2 + 6 + 15 + 32 \\ &\quad {}+ 65 + 126 + 238 = 484 \end{aligned}\text{。}

An element xx is the greater element of a two-element subset exactly once for each smaller element of the set, so xx contributes to the sum once per element below it. Sorting the set as 1,2,3,5,8,13,21,34,1, 2, 3, 5, 8, 13, 21, 34, the sum of the list is 0(1)+1(2)+2(3)+3(5)+4(8)+5(13)+6(21)+7(34)=2+6+15+32+65+126+238=484. \begin{aligned} &0(1) + 1(2) + 2(3) + 3(5) \\ &\quad {}+ 4(8) + 5(13) + 6(21) \\ &\quad {}+ 7(34) \\ &= 2 + 6 + 15 + 32 \\ &\quad {}+ 65 + 126 + 238 = 484. \end{aligned}

4.

已知 log⁡10sin⁡x+log⁡10cos⁡x=−1\log_{10} \sin x + \log_{10} \cos x = -1，且 log⁡10(sin⁡x+cos⁡x)\log_{10}(\sin x + \cos x) =12(log⁡10n−1)= \frac{1}{2}(\log_{10} n - 1)，求 nn。

Given that log⁡10sin⁡x+log⁡10cos⁡x=−1\log_{10} \sin x + \log_{10} \cos x = -1 and that log⁡10(sin⁡x+cos⁡x)\log_{10}(\sin x + \cos x) =12(log⁡10n−1),= \frac{1}{2}(\log_{10} n - 1), find n.n.

答案：12
知识点：对数三角恒等式
难度评级：1990
小提示：

合并对数：第一个等式给出 sin⁡xcos⁡x=110\sin x \cos x = \frac{1}{10}。

Add the logarithms: the first equation gives sin⁡xcos⁡x=110\sin x \cos x = \frac{1}{10}

大提示：

将 sin⁡x+cos⁡x\sin x + \cos x 平方，并使用 sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 来求它。

Square sin⁡x+cos⁡x\sin x + \cos x and use sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 to evaluate it

解答：

第一个等式说明 log⁡10(sin⁡xcos⁡x)=−1\log_{10}(\sin x \cos x) = -1，所以 sin⁡xcos⁡x=110\sin x \cos x = \frac{1}{10}。因此 (sin⁡x+cos⁡x)2=sin⁡2x+cos⁡2x+2sin⁡xcos⁡x=1+210=1210。 \begin{aligned} (\sin x + \cos x)^2 &= \sin^2 x + \cos^2 x \\ &\quad {}+ 2 \sin x \cos x \\ &= 1 + \frac{2}{10} = \frac{12}{10} \end{aligned}\text{。}

取对数得 2log⁡10(sin⁡x+cos⁡x)2\log_{10}(\sin x + \cos x) =log⁡101210= \log_{10} \frac{12}{10} =log⁡1012−1= \log_{10} 12 - 1，所以 log⁡10(sin⁡x+cos⁡x)\log_{10}(\sin x + \cos x) =12(log⁡1012−1)= \frac{1}{2}(\log_{10} 12 - 1)，从而 n=12n = 12。

The first equation says log⁡10(sin⁡xcos⁡x)=−1,\log_{10}(\sin x \cos x) = -1, so sin⁡xcos⁡x=110.\sin x \cos x = \frac{1}{10}. Then (sin⁡x+cos⁡x)2=sin⁡2x+cos⁡2x+2sin⁡xcos⁡x=1+210=1210. \begin{aligned} (\sin x + \cos x)^2 &= \sin^2 x + \cos^2 x \\ &\quad {}+ 2 \sin x \cos x \\ &= 1 + \frac{2}{10} = \frac{12}{10}. \end{aligned}

Taking logarithms, 2log⁡10(sin⁡x+cos⁡x)2\log_{10}(\sin x + \cos x) =log⁡101210= \log_{10} \frac{12}{10} =log⁡1012−1,= \log_{10} 12 - 1, so log⁡10(sin⁡x+cos⁡x)\log_{10}(\sin x + \cos x) =12(log⁡1012−1)= \frac{1}{2}(\log_{10} 12 - 1) and n=12.n = 12.

5.

考虑所有位于一个长、宽、高分别为 33、44、55 个单位的长方体（盒子）内部，或与它的距离不超过一个单位的点。已知这些点所组成的集合体积为 m+nπp\frac{m + n\pi}{p}，其中 mm、nn 和 pp 是正整数，且 nn 与 pp 互质。求 m+n+pm + n + p。

Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures 33 by 44 by 55 units. Given that the volume of this set is m+nπp,\frac{m + n\pi}{p}, where m,m, n,n, and pp are positive integers, and nn and pp are relatively prime, find m+n+p.m + n + p.

答案：505
知识点：体积长方体圆柱球
难度评级：2210
小提示：

这个区域由长方体本身、每个面外侧的薄层、每条棱上的四分之一圆柱，以及每个顶点处的球八分体组成。

The region is the box plus slabs over each face, quarter-cylinders along each edge, and sphere octants at each corner

大提示：

十二个四分之一圆柱合成三个完整圆柱，高分别为 33、44、55；八个球八分体合成一个单位球。

The twelve quarter-cylinders form three full cylinders with heights 3,3, 4,4, 5,5, and the eight octants form one unit sphere

解答：

该区域由长方体本身、从六个面向外伸出的厚度为 11 的薄层、沿十二条棱的半径为 11 的四分之一圆柱，以及八个顶点处半径为 11 的球八分体组成。长方体体积为 3⋅4⋅5=603 \cdot 4 \cdot 5 = 60，各薄层总计 2(3⋅4+3⋅5+4⋅5)=942(3 \cdot 4 + 3 \cdot 5 + 4 \cdot 5) = 94。

与每个维度平行的四条棱上的四分之一圆柱合并成一个完整圆柱，所以这些圆柱总体积为 π⋅12(3+4+5)=12π\pi \cdot 1^2 (3 + 4 + 5) = 12\pi。八个球八分体合并成一个单位球，体积为 4π3\frac{4\pi}{3}。

总体积为 60+94+12π+4π3=154+40π3=462+40π3 \begin{aligned} &60 + 94 + 12\pi \\ &\quad {}+ \frac{4\pi}{3} = 154 + \frac{40\pi}{3} \\ &= \frac{462 + 40\pi}{3} \end{aligned} 因此 m+n+pm + n + p =462+40+3= 462 + 40 + 3 =505= 505。

The region consists of the box itself, six slabs of thickness 11 projecting outward from the faces, quarter-cylinders of radius 11 along the twelve edges, and eighth-spheres of radius 11 at the eight corners. The box has volume 3⋅4⋅5=60,3 \cdot 4 \cdot 5 = 60, and the slabs total 2(3⋅4+3⋅5+4⋅5)=94.2(3 \cdot 4 + 3 \cdot 5 + 4 \cdot 5) = 94.

The four quarter-cylinders along edges parallel to each dimension combine into a full cylinder, so the cylinders total π⋅12(3+4+5)=12π.\pi \cdot 1^2 (3 + 4 + 5) = 12\pi. The eight octants combine into one unit sphere of volume 4π3.\frac{4\pi}{3}.

The total volume is 60+94+12π+4π3=154+40π3=462+40π3, \begin{aligned} &60 + 94 + 12\pi \\ &\quad {}+ \frac{4\pi}{3} = 154 + \frac{40\pi}{3} \\ &= \frac{462 + 40\pi}{3}, \end{aligned} so m+n+pm + n + p =462+40+3= 462 + 40 + 3 =505.= 505.

6.

一个 11 乘 11 乘 11 立方体的八个顶点中，任取三个作为三角形的顶点。所有这类三角形的面积之和为 m+n+pm + \sqrt{n} + \sqrt{p}，其中 mm、nn 和 pp 是整数。求 m+n+pm + n + p。

The sum of the areas of all triangles whose vertices are also vertices of a 11 by 11 by 11 cube is m+n+p,m + \sqrt{n} + \sqrt{p}, where m,m, n,n, and pp are integers. Find m+n+p.m + n + p.

答案：348
知识点：正方体三角形面积分类讨论
难度评级：2370
小提示：

每条边都是立方体的棱、长度为 2\sqrt{2} 的面对角线或长度为 3\sqrt{3} 的体对角线；对 (83)=56\binom{8}{3} = 56 个三角形分类。

Each side is a cube edge, a face diagonal of length 2,\sqrt{2}, or a space diagonal of length 3;\sqrt{3}; classify the (83)=56\binom{8}{3} = 56 triangles

大提示：

三种类型分别由两条棱和一条面对角线、三条面对角线，以及一条棱、一条面对角线和一条体对角线组成；数量分别为 2424、88 和 2424。

The three types are edge-edge-face diagonal, three face diagonals, and edge-face diagonal-space diagonal; there are 24,24, 8,8, and 2424 of them

解答：

这类三角形的每条边都是立方体棱、长度为 2\sqrt{2} 的面对角线，或长度为 3\sqrt{3} 的体对角线。只会出现三种形状。由两条相邻棱和一条面对角线组成的三角形是直角三角形，面积为 12\frac{1}{2}；每个面有 44 个，共 2424 个。由三条面对角线组成的三角形是等边三角形，面积为 32\frac{\sqrt{3}}{2}；立方体的 88 个顶点中，每个顶点相邻的三个顶点都确定一个这样的三角形，所以有 88 个。由一条棱、一条面对角线和一条体对角线组成的三角形是直角三角形，直角边为 11 和 2\sqrt{2}，面积为 22\frac{\sqrt{2}}{2}；44 条体对角线中的每一条，都可与不在该对角线上的 66 个顶点各形成一个，所以有 2424 个。（确实 24+8+24=(83)=5624 + 8 + 24 = \binom{8}{3} = 56。）

总面积为 24⋅12+8⋅32+24⋅22=12+43+122=12+48+288 \begin{aligned} &24 \cdot \frac{1}{2} + 8 \cdot \frac{\sqrt{3}}{2} + 24 \cdot \frac{\sqrt{2}}{2} \\ &= 12 + 4\sqrt{3} + 12\sqrt{2} \\ &= 12 + \sqrt{48} + \sqrt{288} \end{aligned} 因此 m+n+pm + n + p =12+48+288= 12 + 48 + 288 =348= 348。

Every side of such a triangle is a cube edge, a face diagonal of length 2,\sqrt{2}, or a space diagonal of length 3.\sqrt{3}. Only three shapes occur. A triangle of two adjacent edges and a face diagonal is right with area 12;\frac{1}{2}; there are 44 per face, or 24.24. A triangle of three face diagonals is equilateral with area 32;\frac{\sqrt{3}}{2}; each is determined by the three vertices adjacent to one of the 88 cube vertices, so there are 8.8. A triangle of an edge, a face diagonal, and a space diagonal is right with legs 11 and 2,\sqrt{2}, so its area is 22;\frac{\sqrt{2}}{2}; each of the 44 space diagonals forms one with each of the 66 vertices off that diagonal, so there are 24.24. (Indeed 24+8+24=(83)=56.24 + 8 + 24 = \binom{8}{3} = 56.)

The total area is 24⋅12+8⋅32+24⋅22=12+43+122=12+48+288, \begin{aligned} &24 \cdot \frac{1}{2} + 8 \cdot \frac{\sqrt{3}}{2} + 24 \cdot \frac{\sqrt{2}}{2} \\ &= 12 + 4\sqrt{3} + 12\sqrt{2} \\ &= 12 + \sqrt{48} + \sqrt{288}, \end{aligned} so m+n+pm + n + p =12+48+288= 12 + 48 + 288 =348.= 348.

7.

点 BB 在 AC‾\overline{AC} 上，且 AB=9AB = 9、BC=21BC = 21。点 DD 不在 AC‾\overline{AC} 上，并满足 AD=CDAD = CD，且 ADAD 和 BDBD 都是整数。设 ss 为 △ACD\triangle ACD 所有可能周长之和。求 ss。

Point BB is on AC‾\overline{AC} with AB=9AB = 9 and BC=21.BC = 21. Point DD is not on AC‾\overline{AC} so that AD=CD,AD = CD, and ADAD and BDBD are integers. Let ss be the sum of all possible perimeters of △ACD.\triangle ACD. Find s.s.

答案：380
知识点：平方差勾股定理丢番图方程
难度评级：2270
小提示：

从 DD 向 AC‾\overline{AC} 作垂线，其垂足是 AC‾\overline{AC} 的中点；该中点与点 BB 相距 66 个单位。

The foot of the perpendicular from DD to AC‾\overline{AC} is the midpoint of AC‾,\overline{AC}, which is 66 units from BB

大提示：

设 AD=aAD = a、BD=bBD = b，两个直角三角形给出 a2−b2=152−62=189a^2 - b^2 = 15^2 - 6^2 = 189；将它因式分解。

With AD=aAD = a and BD=b,BD = b, the two right triangles give a2−b2=152−62=189;a^2 - b^2 = 15^2 - 6^2 = 189; factor it

解答：

设 AD=CD=aAD = CD = a、BD=bBD = b，并设 EE 是从 DD 到 AC‾\overline{AC} 的垂足。因为 AD=CDAD = CD，点 EE 是 AC‾\overline{AC} 的中点，所以 AE=15AE = 15 且 BE=15−9=6BE = 15 - 9 = 6。直角三角形 DEADEA 与 DEBDEB 共有边 DEDE，因此 a2−152=DE2=b2−62a^2 - 15^2 = DE^2 = b^2 - 6^2 即 (a+b)(a−b)=189。(a+b)(a-b) = 189\text{。}

分解 189=189⋅1189 = 189 \cdot 1 =63⋅3= 63 \cdot 3 =27⋅7=21⋅9= 27 \cdot 7 = 21 \cdot 9，得到 (a,b)=(95,94)(a, b) = (95, 94)、(33,30)(33, 30)、(17,10)(17, 10) 和 (15,6)(15, 6)。最后一组舍去：b=6b = 6 会使 DD 落在 AC‾\overline{AC} 上。每个有效的数对给出周长 2a+302a + 30。

因此 s=(190+30)s = (190 + 30) +(66+30)+ (66 + 30) +(34+30)+ (34 + 30) =220+96+64= 220 + 96 + 64 =380= 380。

Let AD=CD=aAD = CD = a and BD=b,BD = b, and let EE be the foot of the perpendicular from DD to AC‾.\overline{AC}. Since AD=CD,AD = CD, point EE is the midpoint of AC‾,\overline{AC}, so AE=15AE = 15 and BE=15−9=6.BE = 15 - 9 = 6. The right triangles DEADEA and DEBDEB share leg DE,DE, so a2−152=DE2=b2−62,a^2 - 15^2 = DE^2 = b^2 - 6^2, that is (a+b)(a−b)=189.(a+b)(a-b) = 189.

The factorizations 189=189⋅1189 = 189 \cdot 1 =63⋅3= 63 \cdot 3 =27⋅7=21⋅9= 27 \cdot 7 = 21 \cdot 9 give (a,b)=(95,94),(a, b) = (95, 94), (33,30),(33, 30), (17,10),(17, 10), and (15,6).(15, 6). The last is rejected: b=6b = 6 would put DD on AC‾.\overline{AC}. Each valid pair gives a triangle with perimeter 2a+30.2a + 30.

Therefore s=(190+30)s = (190 + 30) +(66+30)+ (66 + 30) +(34+30)+ (34 + 30) =220+96+64= 220 + 96 + 64 =380.= 380.

8.

在一个由四个递增正整数组成的数列中，前三项成等差数列，后三项成等比数列，且第一项与第四项相差 3030。求这四项之和。

In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by 30.30. Find the sum of the four terms.

答案：129
知识点：等差数列等比数列分类讨论
难度评级：2210
小提示：

将四项写成 aa、a+da + d、a+2da + 2d 和 a+30a + 30。

Write the terms as a,a, a+d,a + d, a+2d,a + 2d, and a+30a + 30

大提示：

等比条件化为 3a(10−d)=2d(2d−15)3a(10 - d) = 2d(2d - 15)，因此 10−d10 - d 与 2d−152d - 15 必须同号。

The geometric condition becomes 3a(10−d)=2d(2d−15),3a(10 - d) = 2d(2d - 15), so 10−d10 - d and 2d−152d - 15 must have the same sign

解答：

将四项写成 aa、a+da + d、a+2da + 2d 和 a+30a + 30，其中 aa 和 dd 是正整数。后三项成等比数列给出 (a+30)(a+d)=(a+2d)2(a + 30)(a + d) = (a + 2d)^2。展开并化简得 30a+30d=3ad+4d230a + 30d = 3ad + 4d^2 即 3a(10−d)=2d(2d−15)。3a(10 - d) = 2d(2d - 15)\text{。}

因为 a,d>0a, d \gt 0，因子 10−d10 - d 和 2d−152d - 15 必须同号，故 7.5<d<107.5 \lt d \lt 10，所以 d=8d = 8 或 d=9d = 9。当 d=8d = 8 时，得到 6a=166a = 16，没有整数解。当 d=9d = 9 时，得到 3a=543a = 54，所以 a=18a = 18。

数列为 18,27,36,4818, 27, 36, 48（确实 27,36,4827, 36, 48 的公比为 43\frac{4}{3}），其和为 18+27+36+48=12918 + 27 + 36 + 48 = 129。

Write the terms as a,a, a+d,a + d, a+2d,a + 2d, and a+30,a + 30, where aa and dd are positive integers. The geometric condition on the last three terms says (a+30)(a+d)=(a+2d)2.(a + 30)(a + d) = (a + 2d)^2. Expanding both sides and simplifying, 30a+30d=3ad+4d2,30a + 30d = 3ad + 4d^2, that is 3a(10−d)=2d(2d−15).3a(10 - d) = 2d(2d - 15).

Since a,d>0,a, d \gt 0, the factors 10−d10 - d and 2d−152d - 15 must have the same sign, forcing 7.5<d<10,7.5 \lt d \lt 10, so d=8d = 8 or d=9.d = 9. For d=8,d = 8, we get 6a=16,6a = 16, which has no integer solution. For d=9,d = 9, we get 3a=54,3a = 54, so a=18.a = 18.

The sequence is 18,27,36,4818, 27, 36, 48 (indeed 27,36,4827, 36, 48 has ratio 43\frac{4}{3}), and the sum is 18+27+36+48=129.18 + 27 + 36 + 48 = 129.

9.

一个从 10001000 到 99999999（含两端）的整数，如果它最左边两个数字之和等于最右边两个数字之和，就称为平衡数。有多少个平衡整数？

An integer between 10001000 and 9999,9999, inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits. How many balanced integers are there?

答案：615
知识点：数字分类讨论前n项平方和
难度评级：2430
小提示：

按共同的两位数字和 ss 分组；分别计数最左两位（首位非零）和最右两位

Group by the common digit-pair sum s;s; count leftmost pairs (first digit nonzero) and rightmost pairs separately

大提示：

当 s≤9s \le 9 时，两边的计数分别为 ss 和 s+1s + 1；当 s≥10s \ge 10 时，两者都等于 19−s19 - s。把乘积求和。

For s≤9s \le 9 the counts are ss and s+1;s + 1; for s≥10s \ge 10 both equal 19−s.19 - s. Sum the products.

解答：

按每对数字的共同和 ss 分组，其中 1≤s≤181 \le s \le 18。当 s≤9s \le 9 时，最左两位（第一位至少为 11）可用 ss 种方式形成，最右两位可用 s+1s + 1 种方式形成。当 s≥10s \ge 10 时，每个数位都至少为 s−9s - 9，因此每个数位对都有 19−s19 - s 种方式。

总数为 ∑s=19s(s+1)+∑s=1018(19−s)2=∑s=19(s2+s)+∑k=19k2=2⋅285+45=615。 \begin{aligned} &\sum_{s=1}^{9} s(s+1) + \sum_{s=10}^{18} (19 - s)^2 \\ &= \sum_{s=1}^{9} (s^2 + s) + \sum_{k=1}^{9} k^2 \\ &= 2 \cdot 285 + 45 = 615 \end{aligned}\text{。}

Group the balanced integers by the common sum ss of each digit pair, where 1≤s≤18.1 \le s \le 18. For s≤9,s \le 9, the leftmost pair (first digit at least 11) can be formed in ss ways and the rightmost pair in s+1s + 1 ways. For s≥10,s \ge 10, both digits of each pair must be at least s−9,s - 9, giving 19−s19 - s ways for each pair.

The total count is ∑s=19s(s+1)+∑s=1018(19−s)2=∑s=19(s2+s)+∑k=19k2=2⋅285+45=615. \begin{aligned} &\sum_{s=1}^{9} s(s+1) + \sum_{s=10}^{18} (19 - s)^2 \\ &= \sum_{s=1}^{9} (s^2 + s) + \sum_{k=1}^{9} k^2 \\ &= 2 \cdot 285 + 45 = 615. \end{aligned}

10.

三角形 ABCABC 是等腰三角形，AC=BCAC = BC，且 ∠ACB=106∘\angle ACB = 106^\circ。点 MM 在三角形内部，满足 ∠MAC=7∘\angle MAC = 7^\circ 且 ∠MCA=23∘\angle MCA = 23^\circ。求 ∠CMB\angle CMB 的度数。

Triangle ABCABC is isosceles with AC=BCAC = BC and ∠ACB=106∘.\angle ACB = 106^\circ. Point MM is in the interior of the triangle so that ∠MAC=7∘\angle MAC = 7^\circ and ∠MCA=23∘.\angle MCA = 23^\circ. Find the number of degrees in ∠CMB.\angle CMB.

答案：83
知识点：正弦定理余弦定理等腰三角形
难度评级：2920
小提示：

底角为 37∘37^\circ。在三角形 AMCAMC 中，取 AC=1AC = 1，由正弦定理可得 CM=2sin⁡7∘CM = 2\sin 7^\circ。

The base angles are 37∘.37^\circ. In triangle AMC,AMC, the Law of Sines with AC=1AC = 1 gives CM=2sin⁡7∘.CM = 2\sin 7^\circ.

大提示：

在三角形 BMCBMC 中使用余弦定理，并利用 ∠MCB\angle MCB 的余弦等于 sin⁡7∘\sin 7^\circ 可证明 MB=CBMB = CB。

Apply the Law of Cosines in triangle BMC,BMC, using the fact that the cosine of ∠MCB\angle MCB equals sin⁡7∘,\sin 7^\circ, to show MB=CBMB = CB

解答：

设 AC=BC=1AC = BC = 1。在三角形 AMCAMC 中，AA 与 CC 处的角分别为 7∘7^\circ 和 23∘23^\circ，所以 ∠AMC=150∘\angle AMC = 150^\circ，由正弦定理得 CM=sin⁡7∘sin⁡150∘=2sin⁡7∘。CM = \frac{\sin 7^\circ}{\sin 150^\circ} = 2\sin 7^\circ\text{。}

又 ∠MCB=106∘−23∘=83∘\angle MCB = 106^\circ - 23^\circ = 83^\circ，其余弦为 sin⁡7∘\sin 7^\circ。在三角形 BMCBMC 中用余弦定理：MB2=CM2+CB2−2⋅CM⋅CBcos⁡83∘=4sin⁡27∘+1−4sin⁡27∘=1。 \begin{aligned} MB^2 &= CM^2 + CB^2 \\ &\quad {}- 2 \cdot CM \cdot CB \cos 83^\circ \\ &= 4\sin^2 7^\circ + 1 \\ &\quad {}- 4\sin^2 7^\circ = 1 \end{aligned}\text{。}

因此 MB=1=CBMB = 1 = CB，三角形 BMCBMC 为等腰三角形，∠CMB=∠MCB=83∘\angle CMB = \angle MCB = 83^\circ。答案是 8383。

Assume AC=BC=1.AC = BC = 1. In triangle AMC,AMC, the angles at AA and CC are 7∘7^\circ and 23∘,23^\circ, so ∠AMC=150∘,\angle AMC = 150^\circ, and the Law of Sines gives CM=sin⁡7∘sin⁡150∘=2sin⁡7∘.CM = \frac{\sin 7^\circ}{\sin 150^\circ} = 2\sin 7^\circ.

Also ∠MCB=106∘−23∘=83∘,\angle MCB = 106^\circ - 23^\circ = 83^\circ, whose cosine is sin⁡7∘.\sin 7^\circ. The Law of Cosines in triangle BMCBMC then gives MB2=CM2+CB2−2⋅CM⋅CBcos⁡83∘=4sin⁡27∘+1−4sin⁡27∘=1. \begin{aligned} MB^2 &= CM^2 + CB^2 \\ &\quad {}- 2 \cdot CM \cdot CB \cos 83^\circ \\ &= 4\sin^2 7^\circ + 1 \\ &\quad {}- 4\sin^2 7^\circ = 1. \end{aligned}

So MB=1=CB,MB = 1 = CB, making triangle BMCBMC isosceles with ∠CMB=∠MCB=83∘.\angle CMB = \angle MCB = 83^\circ. The answer is 83.83.

11.

随机选取角 xx，使其位于区间 0∘<x<90∘0^\circ \lt x \lt 90^\circ。设 pp 为 sin⁡2x\sin^2 x、cos⁡2x\cos^2 x 和 sin⁡xcos⁡x\sin x \cos x 不能作为一个三角形三边长的概率。已知 p=dnp = \frac{d}{n}，其中 dd 是 arctan⁡m\arctan m 的度数，且 mm 和 nn 是满足 m+n<1000m + n \lt 1000 的正整数。求 m+nm + n。

An angle xx is chosen at random from the interval 0∘<x<90∘.0^\circ \lt x \lt 90^\circ. Let pp be the probability that the numbers sin⁡2x,\sin^2 x, cos⁡2x,\cos^2 x, and sin⁡xcos⁡x\sin x \cos x are not the lengths of the sides of a triangle. Given that p=dn,p = \frac{d}{n}, where dd is the number of degrees in arctan⁡m\arctan m and mm and nn are positive integers with m+n<1000,m + n \lt 1000, find m+n.m + n.

答案：92
知识点：几何概率三角不等式三角恒等式
难度评级：2710
小提示：

当 0∘<x≤45∘0^\circ \lt x \le 45^\circ 时，三个数中最大的是 cos⁡2x\cos^2 x，因此只有一个三角形不等式可能失败。

For 0∘<x≤45∘0^\circ \lt x \le 45^\circ the largest of the three numbers is cos⁡2x,\cos^2 x, so only one triangle inequality can fail

大提示：

cos⁡2x≥sin⁡2x+sin⁡xcos⁡x\cos^2 x \ge \sin^2 x + \sin x \cos x 可通过倍角公式整理为 tan⁡2x≤2\tan 2x \le 2。

cos⁡2x≥sin⁡2x+sin⁡xcos⁡x\cos^2 x \ge \sin^2 x + \sin x \cos x rearranges via double angles to tan⁡2x≤2\tan 2x \le 2

解答：

将 xx 替换为 90∘−x90^\circ - x 会交换 sin⁡x\sin x 与 cos⁡x\cos x，所以在 (45∘,90∘)(45^\circ, 90^\circ) 上的失败概率与在 (0∘,45∘)(0^\circ, 45^\circ) 上相同，只需考虑 0∘<x≤45∘0^\circ \lt x \le 45^\circ。在此范围内 cos⁡2x≥sin⁡xcos⁡x≥sin⁡2x\cos^2 x \ge \sin x \cos x \ge \sin^2 x，因此这三个数不能组成三角形当且仅当 cos⁡2x≥sin⁡2x+sin⁡xcos⁡x。\cos^2 x \ge \sin^2 x + \sin x \cos x\text{。}

因为 cos⁡2x−sin⁡2x=cos⁡2x\cos^2 x - \sin^2 x = \cos 2x，且 sin⁡xcos⁡x=12sin⁡2x\sin x \cos x = \frac{1}{2}\sin 2x，这等价于 cos⁡2x≥12sin⁡2x\cos 2x \ge \frac{1}{2} \sin 2x，即 tan⁡2x≤2\tan 2x \le 2。由于正切函数在该范围内递增，这恰好在 x≤12arctan⁡2x \le \frac{1}{2}\arctan 2 时发生。

因此 p=12arctan⁡245∘=arctan⁡290∘p = \frac{\frac{1}{2}\arctan 2}{45^\circ} = \frac{\arctan 2}{90^\circ} 所以 m=2m = 2、n=90n = 90，且 m+n=92<1000m + n = 92 \lt 1000，答案为 9292。

Replacing xx by 90∘−x90^\circ - x swaps sin⁡x\sin x and cos⁡x,\cos x, so the failure probability on (45∘,90∘)(45^\circ, 90^\circ) matches that on (0∘,45∘),(0^\circ, 45^\circ), and it suffices to consider 0∘<x≤45∘.0^\circ \lt x \le 45^\circ. There cos⁡2x≥sin⁡xcos⁡x≥sin⁡2x,\cos^2 x \ge \sin x \cos x \ge \sin^2 x, so the three numbers fail to form a triangle exactly when cos⁡2x≥sin⁡2x+sin⁡xcos⁡x.\cos^2 x \ge \sin^2 x + \sin x \cos x.

Since cos⁡2x−sin⁡2x=cos⁡2x\cos^2 x - \sin^2 x = \cos 2x and sin⁡xcos⁡x=12sin⁡2x,\sin x \cos x = \frac{1}{2}\sin 2x, this says cos⁡2x≥12sin⁡2x,\cos 2x \ge \frac{1}{2} \sin 2x, i.e. tan⁡2x≤2.\tan 2x \le 2. Because tangent increases on this range, that happens exactly for x≤12arctan⁡2.x \le \frac{1}{2}\arctan 2.

Therefore p=12arctan⁡245∘=arctan⁡290∘,p = \frac{\frac{1}{2}\arctan 2}{45^\circ} = \frac{\arctan 2}{90^\circ}, so m=2m = 2 and n=90,n = 90, with m+n=92<1000,m + n = 92 \lt 1000, and the answer is 92.92.

12.

在凸四边形 ABCDABCD 中，∠A≅∠C\angle A \cong \angle C，AB=CD=180AB = CD = 180，且 AD≠BCAD \ne BC。四边形 ABCDABCD 的周长为 640640。求 ⌊1000cos⁡A⌋\lfloor 1000 \cos A \rfloor。（记号 ⌊x⌋\lfloor x \rfloor 表示小于或等于 xx 的最大整数。）

In convex quadrilateral ABCD,ABCD, ∠A≅∠C,\angle A \cong \angle C, AB=CD=180,AB = CD = 180, and AD≠BC.AD \ne BC. The perimeter of ABCDABCD is 640.640. Find ⌊1000cos⁡A⌋.\lfloor 1000 \cos A \rfloor. (The notation ⌊x⌋\lfloor x \rfloor means the greatest integer that is less than or equal to x.x.)

答案：777
知识点：余弦定理平方差
难度评级：2560
小提示：

用余弦定理计算 BD2BD^2：分别在三角形 ABDABD 和 CDBCDB 中计算，并令结果相等。

Compute BD2BD^2 by the Law of Cosines in triangles ABDABD and CDB,CDB, and set the results equal

大提示：

因为 AD≠BCAD \ne BC，除以 AD−BCAD - BC 后得到 cos⁡A=AD+BC360\cos A = \frac{AD + BC}{360}，且 AD+BC=640−360AD + BC = 640 - 360。

Because AD≠BC,AD \ne BC, dividing by AD−BCAD - BC leaves cos⁡A=AD+BC360,\cos A = \frac{AD + BC}{360}, and AD+BC=640−360AD + BC = 640 - 360

解答：

设 ∠A=∠C=α\angle A = \angle C = \alpha，AD=xAD = x，BC=yBC = y。对角线 BDBD 在三角形 ABDABD 和 CDBCDB 中分别用余弦定理：BD2=x2+1802−2⋅180xcos⁡α=y2+1802−2⋅180ycos⁡α。 \begin{aligned} BD^2 &= x^2 + 180^2 \\ &\quad {}- 2 \cdot 180x\cos\alpha \\ &= y^2 + 180^2 \\ &\quad {}- 2 \cdot 180y\cos\alpha \end{aligned}\text{。}

整理得 x2−y2=2⋅180(x−y)cos⁡αx^2 - y^2 = 2 \cdot 180(x - y)\cos\alpha，因为 x≠yx \ne y 可除以 x−yx - y：cos⁡α=x+y360=640−2⋅180360=280360=79。 \begin{aligned} \cos\alpha &= \frac{x + y}{360} \\ &= \frac{640 - 2 \cdot 180}{360} \\ &= \frac{280}{360} = \frac{7}{9} \end{aligned}\text{。}

于是 1000cos⁡A=70009=777.7…1000\cos A = \frac{7000}{9} = 777.7\ldots，所以 ⌊1000cos⁡A⌋=777\lfloor 1000\cos A \rfloor = 777。

Let ∠A=∠C=α,\angle A = \angle C = \alpha, AD=x,AD = x, and BC=y.BC = y. Applying the Law of Cosines to diagonal BDBD in triangles ABDABD and CDB,CDB, BD2=x2+1802−2⋅180xcos⁡α=y2+1802−2⋅180ycos⁡α. \begin{aligned} BD^2 &= x^2 + 180^2 \\ &\quad {}- 2 \cdot 180x\cos\alpha \\ &= y^2 + 180^2 \\ &\quad {}- 2 \cdot 180y\cos\alpha. \end{aligned}

Rearranging gives x2−y2=2⋅180(x−y)cos⁡α,x^2 - y^2 = 2 \cdot 180(x - y)\cos\alpha, and since x≠yx \ne y we may divide by x−y:x - y: cos⁡α=x+y360=640−2⋅180360=280360=79. \begin{aligned} \cos\alpha &= \frac{x + y}{360} \\ &= \frac{640 - 2 \cdot 180}{360} \\ &= \frac{280}{360} = \frac{7}{9}. \end{aligned}

Then 1000cos⁡A=70009=777.7…,1000\cos A = \frac{7000}{9} = 777.7\ldots, so ⌊1000cos⁡A⌋=777.\lfloor 1000\cos A \rfloor = 777.

13.

设 NN 为不超过 20032003 的正整数中，其 22 进制表示里 11 的个数多于 00 的个数的整数个数。求 NN 除以 10001000 的余数。

Let NN be the number of positive integers that are less than or equal to 20032003 and whose base-22 representation has more 11’s than 00’s. Find the remainder when NN is divided by 1000.1000.

答案：155
知识点：进制组合杨辉三角对称性
难度评级：2920
小提示：

每个这样的整数最多有 1111 个二进制位；对有 (d+1)(d+1) 个二进制位的数，按数字 11 的个数用帕斯卡三角形第 dd 行计数。

Every such integer has at most 1111 binary digits; for (d+1)(d+1)-digit numbers, count by the number of 11’s using row dd of Pascal’s triangle

大提示：

由对称性，第 00 到第 1010 行贡献 2047+3512\frac{2047 + 351}{2} 个数；再去掉从 20042004 到 20472047 中符合条件的整数。

Rows 00 through 1010 contribute 2047+3512\frac{2047 + 351}{2} numbers by symmetry; then discard the qualifying integers from 20042004 to 2047.2047.

解答：

因为 2003<211=20482003 \lt 2^{11} = 2048，所考虑的每个整数最多有 1111 个二进制位。一个有 (d+1)(d+1) 个二进制位的数以 11 开头；若再选 kk 个 11，选择范围是余下的 dd 位，便可得到 (dk)\binom{d}{k} 个含有 k+1k + 1 个一的数。数字 11 的个数多于数字 00 的个数，恰好发生在 k≥d2k \ge \frac{d}{2} 时。因此，所有不超过 20472047 的数的计数就是帕斯卡三角形第 00 到第 1010 行中，中间项及其右侧各项之和。

这些行的总和为 1+2+⋯+210=20471 + 2 + \cdots + 2^{10} = 2047，中间项之和为 ∑i=05(2ii)\sum_{i=0}^{5}\binom{2i}{i} =1+2+6+20+70+252= 1 + 2 + 6 + 20 + 70 + 252 =351= 351，所以由对称性，计数为 2047+3512=1199\frac{2047 + 351}{2} = 1199。

这 4444 个从 20042004 到 20472047 的整数都大于 1984=1111100000021984 = 11111000000_2，因此每个都有前缀 1111111111 并且至少还多一个 11，所以在十一位中至少有六个 11，这 4444 个全都被计入。因此 N=1199−44=1155N = 1199 - 44 = 1155，除以 10001000 的余数为 155155。

Since 2003<211=2048,2003 \lt 2^{11} = 2048, every integer in question has at most 1111 binary digits. A (d+1)(d+1)-digit binary number starts with 1,1, and choosing kk more 11’s among the remaining dd digits gives (dk)\binom{d}{k} numbers with k+1k + 1 ones; the 11’s outnumber the 00’s exactly when k≥d2.k \ge \frac{d}{2}. So the count over all numbers up to 20472047 is the total of the entries on or to the right of the center of rows 00 through 1010 of Pascal’s triangle.

Those rows sum to 1+2+⋯+210=2047,1 + 2 + \cdots + 2^{10} = 2047, and the central entries sum to ∑i=05(2ii)\sum_{i=0}^{5}\binom{2i}{i} =1+2+6+20+70+252= 1 + 2 + 6 + 20 + 70 + 252 =351,= 351, so by symmetry the count is 2047+3512=1199.\frac{2047 + 351}{2} = 1199.

The 4444 integers from 20042004 to 20472047 all exceed 1984=111110000002,1984 = 11111000000_2, so each has the prefix 1111111111 plus at least one more 1,1, hence at least six 11’s among eleven digits — all 4444 were counted. Therefore N=1199−44=1155,N = 1199 - 44 = 1155, whose remainder upon division by 10001000 is 155.155.

14.

分数 mn\frac{m}{n} 中，mm 和 nn 是互质正整数且 m<nm \lt n；它的十进制表示中依次连续出现数字 22、55 和 11。求满足条件的最小 nn。

The decimal representation of mn,\frac{m}{n}, where mm and nn are relatively prime positive integers and m<n,m \lt n, contains the digits 2,2, 5,5, and 11 consecutively, and in that order. Find the smallest value of nn for which this is possible.

答案：127
知识点：小数丢番图方程极限情形界定
难度评级：3270
小提示：

说明只需让 251251 紧跟在小数点后出现，因此 2511000≤mn<2521000\frac{251}{1000} \le \frac{m}{n} \lt \frac{252}{1000}。

Show it suffices for 251251 to appear immediately after the decimal point, so 2511000≤mn<2521000\frac{251}{1000} \le \frac{m}{n} \lt \frac{252}{1000}

大提示：

利用相邻分数 14<32127<63250\frac14 \lt \frac{32}{127} \lt \frac{63}{250}：每对相邻分数的交叉差都是 11，这可给出夹在它们之间的分数的分母下界。

Use the neighboring fractions 14<32127<63250:\frac14 \lt \frac{32}{127} \lt \frac{63}{250}: each adjacent pair has cross-difference 1,1, which bounds the denominators between them

解答：

只需考虑让 251251 紧接在小数点后出现。事实上，若 mn=0.A251…\frac{m}{n} = 0.A251\ldots，其中 AA 是长度为 k≥1k \ge 1 的数字块，那么 10kmn−A=0.251…10^k \frac{m}{n} - A = 0.251\ldots 是介于 00 与 11 之间的分数，约分后的分母不超过 nn。所以要寻找最小的 nn，使某个 mm 满足 2511000≤mn<2521000\frac{251}{1000} \le \frac{m}{n} \lt \frac{252}{1000} 即 0≤1000m−251n<n。0 \le 1000m - 251n \lt n\text{。}

分数 32127\frac{32}{127} 位于这个区间内，因为 2511000<32127<63250=2521000。 \frac{251}{1000} \lt \frac{32}{127} \lt \frac{63}{250} = \frac{252}{1000}\text{。} 还需证明更小的分母都不可能。使用如下基本事实：若 ab<uv<cd\frac{a}{b} \lt \frac{u}{v} \lt \frac{c}{d}，且 bc−ad=1bc-ad=1，那么 v=b(cv−du)+d(bu−av)≥b+d， \begin{aligned} v &= b(cv-du)+d(bu-av) \\ &\ge b+d\text{，} \end{aligned} 因为括号中的两个量都是正整数。

现在 4⋅32−1⋅127=14\cdot32-1\cdot127=1，且 127⋅63−32⋅250=1127\cdot63-32\cdot250=1。因此，严格介于 14\frac14 与 32127\frac{32}{127} 之间的每个分数，其分母至少为 4+127=1314+127=131；严格介于 32127\frac{32}{127} 与 63250\frac{63}{250} 之间的每个分数，其分母至少为 127+250=377127+250=377。目标区间包含在 (14,63250)\left(\frac14,\frac{63}{250}\right) 内，并包含 32127\frac{32}{127}，所以其中没有分母小于 127127 的分数。

最小可能的 nn 是 127127。

It suffices to make 251251 appear immediately after the decimal point: if mn=0.A251…\frac{m}{n} = 0.A251\ldots with AA a block of k≥1k \ge 1 digits, then 10kmn−A=0.251…10^k \frac{m}{n} - A = 0.251\ldots is a fraction between 00 and 11 whose reduced denominator is at most n.n. So we need the smallest nn admitting an mm with 2511000≤mn<2521000,\frac{251}{1000} \le \frac{m}{n} \lt \frac{252}{1000}, that is 0≤1000m−251n<n.0 \le 1000m - 251n \lt n.

The fraction 32127\frac{32}{127} lies in this interval because 2511000<32127<63250=2521000. \frac{251}{1000} \lt \frac{32}{127} \lt \frac{63}{250} = \frac{252}{1000}. It remains to prove that no smaller denominator works. We use the following elementary fact: if ab<uv<cd\frac{a}{b} \lt \frac{u}{v} \lt \frac{c}{d} and bc−ad=1,bc-ad=1, then v=b(cv−du)+d(bu−av)≥b+d, \begin{aligned} v &= b(cv-du)+d(bu-av) \\ &\ge b+d, \end{aligned} because both parenthesized quantities are positive integers.

Now 4⋅32−1⋅127=14\cdot32-1\cdot127=1 and 127⋅63−32⋅250=1.127\cdot63-32\cdot250=1. Therefore every fraction strictly between 14\frac14 and 32127\frac{32}{127} has denominator at least 4+127=131,4+127=131, while every fraction strictly between 32127\frac{32}{127} and 63250\frac{63}{250} has denominator at least 127+250=377.127+250=377. Since our target interval lies inside (14,63250)\left(\frac14,\frac{63}{250}\right) and contains 32127,\frac{32}{127}, no fraction in it has denominator below 127.127.

The smallest possible value of nn is 127.127.

15.

在 △ABC\triangle ABC 中，AB=360AB = 360，BC=507BC = 507，且 CA=780CA = 780。设 MM 为 CA‾\overline{CA} 的中点，DD 为 CA‾\overline{CA} 上使得 BD‾\overline{BD} 平分角 ABCABC 的点。设 FF 为 BC‾\overline{BC} 上满足 DF‾⊥BD‾\overline{DF} \perp \overline{BD} 的点。若 DF‾\overline{DF} 与 BM‾\overline{BM} 交于 EE，比值 DE:EFDE : EF 可写成 mn\frac{m}{n}，其中 mm 和 nn 是互质正整数。求 m+nm + n。

In △ABC,\triangle ABC, AB=360,AB = 360, BC=507,BC = 507, and CA=780.CA = 780. Let MM be the midpoint of CA‾,\overline{CA}, and let DD be the point on CA‾\overline{CA} such that BD‾\overline{BD} bisects angle ABC.ABC. Let FF be the point on BC‾\overline{BC} such that DF‾⊥BD‾.\overline{DF} \perp \overline{BD}. Suppose that DF‾\overline{DF} meets BM‾\overline{BM} at E.E. The ratio DE:EFDE : EF can 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：289
知识点：角平分线定理平行线相似
难度评级：3370
小提示：

将 FD‾\overline{FD} 过 DD 延长，与射线 BABA 交于 GG。在三角形 BGFBGF 中，BD‾\overline{BD} 既是高又是角平分线，所以 BG=BFBG = BF。

Extend FD‾\overline{FD} past DD to hit ray BABA at G.G. In triangle BGF,BGF, BD‾\overline{BD} is both an altitude and an angle bisector, so BG=BF.BG = BF.

大提示：

取 F1F_1 为 AC‾\overline{AC} 上的点，使 FF1‾∥BM‾\overline{FF_1} \parallel \overline{BM}；则由相似三角形，DE:EF=DM:MF1DE : EF = DM : MF_1。

Take F1F_1 on AC‾\overline{AC} with FF1‾∥BM‾;\overline{FF_1} \parallel \overline{BM}; then DE:EF=DM:MF1DE : EF = DM : MF_1 by similar triangles

解答：

记 c=AB=360c = AB = 360、a=BC=507a = BC = 507、b=CA=780b = CA = 780。将 FD‾\overline{FD} 过 DD 延长，与射线 BABA 在 AA 外侧交于 GG。在三角形 BGFBGF 中，线段 BD‾\overline{BD} 既是角平分线又是高，所以 BG=BF=tBG = BF = t。角平分线还给出 CDDA=ac\frac{CD}{DA} = \frac{a}{c}，因此将梅涅劳斯定理用于直线 GDFGDF 与三角形 ABCABC： AGGB⋅BFFC⋅CDDA=t−ct⋅ta−t⋅ac=1 \begin{aligned} &\frac{AG}{GB} \cdot \frac{BF}{FC} \cdot \frac{CD}{DA} \\ &= \frac{t - c}{t} \cdot \frac{t}{a - t} \cdot \frac{a}{c} = 1 \end{aligned} 所以 t=2aca+c。t = \frac{2ac}{a + c}\text{。}

现在取 F1F_1 为 AC‾\overline{AC} 上满足 FF1‾∥BM‾\overline{FF_1} \parallel \overline{BM} 的点。因为 EE 在 BM‾\overline{BM} 上，有 EM‾∥FF1‾\overline{EM} \parallel \overline{FF_1}，所以三角形 DEMDEM 与 DFF1DFF_1 相似，且 DEEF=DMMF1\frac{DE}{EF} = \frac{DM}{MF_1}。角平分线比例给出 AD=bca+cAD = \frac{bc}{a + c}，因此 DM=b2−bca+c=b(a−c)2(a+c)DM = \frac{b}{2} - \frac{bc}{a+c} = \frac{b(a - c)}{2(a + c)}。又 CF=a−t=a(a−c)a+cCF = a - t = \frac{a(a - c)}{a + c}，所以 CF1=CM⋅CFCB=b2⋅a−ca+cCF_1 = CM \cdot \frac{CF}{CB} = \frac{b}{2} \cdot \frac{a - c}{a + c}，并且 MF1=b2(1−a−ca+c)=bca+cMF_1 = \frac{b}{2}\left(1 - \frac{a - c}{a + c}\right) = \frac{bc}{a + c}。

因此 DEEF=DMMF1=a−c2c=147720=49240 \begin{aligned} \frac{DE}{EF} &= \frac{DM}{MF_1} = \frac{a - c}{2c} \\ &= \frac{147}{720} = \frac{49}{240} \end{aligned} 所以 m+n=49+240=289m + n = 49 + 240 = 289。

Write c=AB=360,c = AB = 360, a=BC=507,a = BC = 507, b=CA=780.b = CA = 780. Extend FD‾\overline{FD} beyond DD to meet ray BABA beyond AA at G.G. In triangle BGF,BGF, segment BD‾\overline{BD} is both an angle bisector and an altitude, so BG=BF=t.BG = BF = t. The bisector also gives CDDA=ac,\frac{CD}{DA} = \frac{a}{c}, so Menelaus’ theorem for line GDFGDF crossing triangle ABCABC says AGGB⋅BFFC⋅CDDA=t−ct⋅ta−t⋅ac=1, \begin{aligned} &\frac{AG}{GB} \cdot \frac{BF}{FC} \cdot \frac{CD}{DA} \\ &= \frac{t - c}{t} \cdot \frac{t}{a - t} \cdot \frac{a}{c} = 1, \end{aligned} so t=2aca+c.t = \frac{2ac}{a + c}.

Now let F1F_1 be the point on AC‾\overline{AC} with FF1‾∥BM‾.\overline{FF_1} \parallel \overline{BM}. Since EE lies on BM‾,\overline{BM}, we have EM‾∥FF1‾,\overline{EM} \parallel \overline{FF_1}, so triangles DEMDEM and DFF1DFF_1 are similar and DEEF=DMMF1.\frac{DE}{EF} = \frac{DM}{MF_1}. The bisector ratio gives AD=bca+c,AD = \frac{bc}{a + c}, so DM=b2−bca+c=b(a−c)2(a+c).DM = \frac{b}{2} - \frac{bc}{a+c} = \frac{b(a - c)}{2(a + c)}. Also CF=a−t=a(a−c)a+c,CF = a - t = \frac{a(a - c)}{a + c}, so CF1=CM⋅CFCB=b2⋅a−ca+cCF_1 = CM \cdot \frac{CF}{CB} = \frac{b}{2} \cdot \frac{a - c}{a + c} and MF1=b2(1−a−ca+c)=bca+c.MF_1 = \frac{b}{2}\left(1 - \frac{a - c}{a + c}\right) = \frac{bc}{a + c}.

Therefore DEEF=DMMF1=a−c2c=147720=49240, \begin{aligned} \frac{DE}{EF} &= \frac{DM}{MF_1} = \frac{a - c}{2c} \\ &= \frac{147}{720} = \frac{49}{240}, \end{aligned} and m+n=49+240=289.m + n = 49 + 240 = 289.