Math Is Fun Forum

oh...my English is poor..so i can not express what i think sometimes.:D but i'm very sure my answer is right.:P

on the image:

O-ABCDE is a part of a Icosahedron.
OA=OB=OC=OD=OE=AB=BC=CD=DE=EA
Since ABCDE is a Regular Pentagon,therefore <EAB=108°.

triangle OAB, OBC, OCD, ODE, OEF are Regular triangles.
let OA=OB=OC=OD=OE=AB=BC=CD=DE=EA=1,
    OF=AF=0.5
then
    BF=EF=sqrt(AB^2-AF^2)=sqrt(3)/2
in triangle EAB

    BE^2=EA^2+AB^2-2*EA*AB*Cos<EAB=2-2Cos108°
in triangle EFB

BE^2=EF^2+FB^2-2*EF*FB*Cos<EFB=3/2-3/2*Cos<EFB
therefore
2-2Cos108°=3/2-3/2*Cos<EFB
<EFB=138.19° 

DO you have msn? i can solve some problems.

en..Draw a straight line AB through the circle at random, and bisect it at the point I. Draw DI from I at right angles to AB, and draw it through to C. Bisect CD at O. point O is the center.:)
or
Draw a straight line AB through the circle at random, and bisect it at the point I.Draw a straight line EF through the circle at random, and bisect it at the point J. Draw a straight line CID at right angles to AB.Draw a straight line  GJH at right angles to EF.
point O is the center.:D

what is "the help of 'Image upload'"?