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《数学通报》2012年第3期第2042号数学问题为:四面体的四个面都为全等的等腰三角形,求等腰三角形底角的范围.原解答首先作四面体S-ABC(如上图),四面均为全等的等腰三角形,其中底边SA、BC长为a、底角为θ,取BC的中点M,连接AM、SM.过S作SO⊥面ABC,垂足为O.易证O在AM上.这是原文讨论的一个基本出发点,也是导致原解
Mathematical Bulletin, 2012, No. 3, No. 2042, mathematics problem is that the four sides of a tetrahedron are all equal isosceles triangles, and the range of base angles of isosceles triangles is sought. The original solution was first made tetrahedral S-ABC ( As shown in the figure above, all sides are equal isosceles triangles, where the bottom SA, BC are a long and the bottom angle is θ. Take the midpoint M of BC, connect AM, SM. The O. Evidence O is on the AM. This is a basic starting point for the original discussion, and it also leads to the original solution.