2016 AIME II 第 7 题

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7.

正方形 ABCDABCDEFGHEFGH 有共同的中心,且 ABEF\overline{AB} \parallel \overline{EF}ABCDABCD 的面积为 20162016EFGHEFGH 的面积是一个较小的正整数。构造正方形 IJKLIJKL,使它的每个顶点都在 ABCDABCD 的一条边上,并且 EFGHEFGH 的每个顶点都在 IJKLIJKL 的一条边上。求 IJKLIJKL 面积的所有可能整数值中,最大值与最小值的差。

Squares ABCDABCD and EFGHEFGH have a common center and ABEF.\overline{AB} \parallel \overline{EF}. The area of ABCDABCD is 2016,2016, and the area of EFGHEFGH is a smaller positive integer. Square IJKLIJKL is constructed so that each of its vertices lies on a side of ABCDABCD and each vertex of EFGHEFGH lies on a side of IJKL.IJKL. Find the difference between the largest and smallest possible integer values for the area of IJKL.IJKL.

答案:840
知识点:正方形(几何)三角学等比数列整除性
难度评级:2920
解答:

若一个边长为 tt 的正方形,其顶点在一个同心、边长为 ss 的正方形边上,并与外正方形成角 θ\theta,则外正方形的每条边被分成长度 tcosθt\cos\thetatsinθt\sin\theta 的两段,所以 s=t(cosθ+sinθ)s = t(\cos\theta + \sin\theta)。这适用于 ABCDABCD(边长 ss)中的 IJKLIJKL(边长 tt),对应某个角 θ\theta。由于 EFAB\overline{EF} \parallel \overline{AB},正方形 EFGHEFGH(边长 uu)与 IJKLIJKL 也成同一个角 θ\theta,所以还有 t=u(cosθ+sinθ)t = u(\cos\theta + \sin\theta)

因此 st=tu\frac{s}{t} = \frac{t}{u},所以三个面积成等比数列:若 TTIJKLIJKL 的面积,则 EFGHEFGH 的面积为 T22016\frac{T^2}{2016}。当 θ\theta(0,90)\left(0^\circ, 90^\circ\right) 中变化时,因子 (cosθ+sinθ)2(\cos\theta + \sin\theta)^2 取遍 (1,2](1, 2] 中的每个值,所以 T=2016(cosθ+sinθ)2T = \frac{2016}{(\cos\theta + \sin\theta)^2} 取遍 [1008,2016)[1008, 2016) 中的每个值(θ=0\theta = 0^\circ 被排除,因为 EFGHEFGH 小于 ABCDABCD)。为了使 T22016\frac{T^2}{2016} 是整数,2016=253272016 = 2^5 \cdot 3^2 \cdot 7 必须整除 T2T^2,这迫使 2337=1682^3 \cdot 3 \cdot 7 = 168 整除 TT

[1008,2016)[1008, 2016)168168 的倍数从 1008100818481848,并且每个都能由合适的 θ\theta 取得,此时 EFGHEFGH 的面积是小于 20162016 的正整数。差为 18481008=8401848 - 1008 = 840

If a square of side tt has its vertices on the sides of a concentric square of side ss and is tilted by angle θ,\theta, each side of the outer square is split into pieces tcosθt\cos\theta and tsinθ,t\sin\theta, so s=t(cosθ+sinθ).s = t(\cos\theta + \sin\theta). This applies to IJKLIJKL (side tt) in ABCDABCD (side ss) with some angle θ.\theta. Since EFAB,\overline{EF} \parallel \overline{AB}, square EFGHEFGH (side uu) makes the same angle θ\theta with IJKL,IJKL, so also t=u(cosθ+sinθ).t = u(\cos\theta + \sin\theta).

Hence st=tu,\frac{s}{t} = \frac{t}{u}, so the three areas form a geometric progression: the area of EFGHEFGH equals T22016,\frac{T^2}{2016}, where TT is the area of IJKL.IJKL. As θ\theta ranges over (0,90),\left(0^\circ, 90^\circ\right), the factor (cosθ+sinθ)2(\cos\theta + \sin\theta)^2 takes every value in (1,2],(1, 2], so T=2016(cosθ+sinθ)2T = \frac{2016}{(\cos\theta + \sin\theta)^2} takes every value in [1008,2016)[1008, 2016) (θ=0\theta = 0^\circ is excluded because EFGHEFGH is smaller than ABCDABCD). For T22016\frac{T^2}{2016} to be an integer, 2016=253272016 = 2^5 \cdot 3^2 \cdot 7 must divide T2,T^2, which forces 2337=1682^3 \cdot 3 \cdot 7 = 168 to divide T.T.

The multiples of 168168 in [1008,2016)[1008, 2016) run from 10081008 to 1848,1848, and each is attained by an appropriate θ,\theta, with the area of EFGHEFGH then a positive integer less than 2016.2016. The difference is 18481008=840.1848 - 1008 = 840.

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