I. AD=BC=C'D, AE=CF=C'F, DE=BE=DF
II.正5边形中,MN=NP=PQ=QD=DM=m, AD=BC=C'D=b, AB=CD=a
∠DMN=∠MNP=∠NPQ=∠PQD=∠QDM=108°
① 由折叠过程,DE=BE,∠ADE=∠CDF=∠QDM-∠ADC=108°-90°=18°
在Rt△ADE中,有AE/AD=tan∠ADE,即AE=AD*tan∠ADE = b*tan18°
a²-b²=(AE+EB)²-AD²
=(AE+ED)²-AD²
=AE²+2AE*ED+ED²-AD²
=2AE²+2AE*ED
=2AE(AE+ED)
=2AE(AE+EB)
=2*b tan18°*a
=2ab tan18°
② 连DN,则∠NDG=∠MDG-∠MDN=∠QDM/2-(180°-∠DMN)/2=18°
GN = NP/2 = m/2,DG = BG = BD/2 = √(a²+b²)/2
在Rt△DNG中,有tan∠NDG = NG/DG = (NP/2) / (BD/2) = NP/BD
即 m = NP = BD tan∠NDG = √(a²+b²) tan18°
③ 连BN.
AE与A'E'关于MN对称,AE交A'E'于点B,因此B在对称轴MN上,即BMN共线
在Rt△ABM中,∠ABM=90°-∠AMB=90°-(180°-108°)=18°
tan∠ABM = AM/AB = (AD-DM)/AB
b = AD
= DM + AB tan∠ABM
= m + a tan18°
④ 作 NH⊥MQ于H,则∠MNH = ∠MNP-∠PNH = 108°-90°=18°
在Rt△MNH中,tan∠MNH = MH/NH
A'H = NG = NP/2 = m/2
有MH = MN sin∠MNH = m sin18°
b = AD
= DM + AM
= DM + A'M
= DM + A'H + MH
= m + m/2 + m sin18°
= (3/2)m + m sin18°
< (3/2)m + m tan18°
因此 ①②③成立,④不成立