Question

正方形ABCD的边长为13,平面ABCD外一点P到正方形各顶点的距离都是13,M,N分别是PA,D

Answer 1

AM=8, BD=13sqrt(2),
ABN用余弦定理
AN^2 = AB^2 + BN^2 - 2AB*BNcos45 = 13^2 + 50 - 2*13*5sqrt(2)/sqrt(2) = 219 - 130 = 89
cosPDB=1/2 DB/PD = sqrt(2)/2
PN^2 = PD^2 + DN^2 - 2PD*DNcosPDB = 13^2 + 128 - 2*13*8sqrt(2)/sqrt(2) = 89
PN=AN = sqrt(89)
cosPAN = 1/2 PA/AN = 13/(2sqrt(89))
三角形MAN应用余弦定理
MN^2 = AM^2 + AN^2 - 2AM*AN*cosPAN
= 64 + 89 - 2*8*sqrt(89)*13/2/sqrt(89) = 64 + 89 - 13*8 = 49
MN = 7