2013年“希望杯”全国数学邀请赛初二第2试第16题为:如图1,点P在正方形ABCD内,△PBC是正三角形,若△BPD的面积是槡3-1,则正方形ABCD的边长是.一、对题目条件的探究图2设PC交BD于点E如图2,由四边形ABCD是正方形,△PBC是正三角形,可以得到∠DCP=30°,PC=CD,从而可知∠CPD=∠CDP=75°,又∠BDC=45°,所以得到∠PDE=30°,即△PDE∽△PCD,也即DP=DE.此外还有∠ABP=30°,∠PBE=15°.所以对条件探究后我 As shown in Figure 1, point P is within a square ABCD, △ PBC is a regular triangle, and if the area of ​​△ BPD is 槡 3-1, then the square ABCD side length is 1. First, the conditions of the topic of exploration Figure 2 Let PC cross BD at point E in Figure 2, the quadrilateral ABCD is a square, △ PBC is a regular triangle, you can get ∠ DCP = 30 °, PC = CD, and thus We can see that ∠CPD = ∠CDP = 75 ° and ∠BDC = 45 °, so we get ∠PDE = 30 °, that is, PDPDE∽ΔPCD, that is, DP = DE. In addition there are ∠ABP = 30 °, ∠PBE = 15 °. So for the conditions to explore me