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题目已知,如图1,在矩形ABCD中,AD=4,CD=3,限定点E在边AB上,点F在边BC上,将△BEF沿EF翻折后压平,落在△EB′F的位置,点B落在形内点B′处,则点B′距点A的最小距离是.文[1]王老师通过分类讨论的方法,最终得到AB′的最小值为1,本文尝试用不等式的最值解决此题.图1证明如图1,连接AF,AB′,设BF=b,依题有0≤b≤4,B′F=b,AF=AB2槡+BF2=槡9+b2.在△AB′F中,AB′>AF-B′F=槡9+b2-b=(槡9+b2-b)(槡9+b2+b)槡9+b2+b=
The subject is known, as shown in Figure 1, in a rectangular ABCD, AD = 4, CD = 3, defining the point E on the side AB, point F on the side BC, folding the ΔBEF along the EF, flattening it, EB’F position, the point B falls within the shape of the point B ’, then the point B’ from the minimum distance A is A. Wen [1] through the classification of discussion, the final result AB ’minimum 1 In this paper we try to solve the problem by using the inequality of the maximum value. Figure 1 shows the connection of AF, AB ’, BF = b in Figure 1, 0≤b≤4, B’F = b, AF = AB2 槡 + BF2 = 槡 9 + b2. In Δ AB’F, AB ’> AF-B’F = 槡 9 + b2-b = 槡 9 + b2- b 槡 9 + b2 + b 槡 9 + b2 + b =