[论文解读] Triangle Tiling IV: A non-isosceles tile with a 120 degree angle

Michael Beeson|arXiv (Cornell University)|Jun 9, 2012

Mathematics and Applications参考文献 15被引用 4

一句话总结

本文研究非等腰三角形密铺，其中密铺块具有120°角，且不与被密铺的三角形ABC相似。通过代数数论、线性代数和几何约束，证明若此类N-密铺存在，则密铺块的最小角不能是π的有理倍数，且密铺块的边长必须为整数，满足c² = a² + b² + ab，且边长之间至少存在一个线性边关系。

ABSTRACT

An N-tiling of triangle ABC by triangle T is a way of writing ABC as a union of N trianglescongruent to T, overlapping only at their boundaries. The triangle T is the tile. The tile may or may not be similar to ABC. We wish to understand possible tilings by completely characterizing the triples (ABC, T, N) such that ABC can be N-tiled by T. In particular, this understanding should enable us to specify for which N there exists a tile T and a triangle ABC that is N-tiled by T; or given N, determine which tiles and triangles can be used for N-tilings; or given ABC, to determine which tiles and N can be used to N-tile ABC. This is one of four papers on this subject. In this paper, we take up the last remaining case: when ABC is not similar to T, and T has a 120 degree angle, and T is not isosceles (although ABC can be isosceles or even equilateral). Here is our result: If there is such an N-tiling, then the smallest angle of the tile is not a rational multiple of \pi. In total there are six tiles with vertices at the vertices of ABC. If the sides of the tile are (a,b,c), then there must be at least one edge relation of the form jb = ua + vc or ja = ub + vc, with j, u, and v all positive. The ratios a/c and b/c are rational, so that after rescaling we can assume the tile has integer sides, which by virtue of the law of cosines satisfy c^2 = a^2 + b^2 + ab. A simple unsolved specific case is when ABC is equilateral and (a,b,c) = (3,5,7). The techniques used in this paper, for the reduction to the integer-sides case, involve linear algebra, elementary field theory and algebraic number theory, as well as geometrical arguments. Quite different methods are required when the sides of the tile are all integers.

研究动机与目标

• 全面表征三元组(ABC, T, N)，其中三角形ABC被一个非等腰的密铺块T以120°角进行N-密铺。
• 确定当ABC与T不相似时，此类密铺存在的N值范围。
• 识别密铺块边长与角度的必要条件，特别关注角度的有理性与边关系。
• 解决该系列三角形密铺论文中最后一个未解问题：非相似、非等腰的密铺块具有120°角。

提出的方法

• 通过缩放与余弦定律将密铺块简化为整数边长，导出方程c² = a² + b² + ab。
• 应用初等域论与代数数论分析角度的有理性与几何约束。
• 利用线性代数建模密铺块边长之间的边关系，要求至少存在一个形如jb = ua + vc或ja = ub + vc的关系，其中j, u, v为正整数。
• 通过几何分析表明，恰好有六个密铺块在ABC的顶点处有顶点。
• 将代数约束与几何密铺构型相结合，排除不可能的情形。
• 系统性地检查边关系与边长比值，推导出N-密铺存在的必要条件。

实验结果

研究问题

• RQ1 对于哪些整数N，存在一个三角形ABC与一个非等腰的密铺块T（具有120°角），使得ABC被T以N-密铺？
• RQ2 此类N-密铺可能存在的条件下，密铺块边长(a, b, c)必须满足什么条件？
• RQ3 在此类密铺中，密铺块的最小角能否是π的有理倍数？
• RQ4 像jb = ua + vc这样的线性边关系在促成或禁止此类密铺中起什么作用？
• RQ5 能否用边长为(3, 5, 7)且具有120°角的密铺块对一个等边三角形ABC进行N-密铺？

主要发现

• 若存在一个非等腰密铺块T（具有120°角）且ABC不与T相似的N-密铺，则T的最小角不能是π的有理倍数。
• 经缩放后，密铺块的边长必须为整数，且满足丢番图方程c² = a² + b² + ab。
• 至少存在一个形如jb = ua + vc或ja = ub + vc的边关系，其中j, u, v为正整数。
• 比值a/c与b/c为有理数，因此密铺块可被缩放为整数边长。
• 在密铺中，恰好有六个密铺块在三角形ABC的顶点处有顶点。
• 等边三角形ABC被(3,5,7)密铺块（具有120°角）密铺的情况仍为开放问题，尽管其满足推导出的条件。