1)triangles AOF and AOD are congruent the bisector of angle DAF is also the bisector of the central angle DOF,M midpoint of arc DF lies on this bisector M lies on the bisector of inscribed angle AFD whose leg AF is tangent to the incercle the bisectors of triangle ADF meet at M=O1 2)in triangle AOF sin(A/2)=r/AO=r'/(AO-r) r'=r(1-sin(A/2)) the same way r''=r(1-sin(B/2);r'''=r(1-sin(C/2)) in triangle DEF,r is the circumradius and r4 is the inradius r4=4rsin((A+B)/4)sin((B+C)/4)sin((C+A)/4) after substitution and appropriate transformation r'+r''+r'''+r4=2r .-.
About the solution of problem 209 by Anonymous - December 31, 2009: How can I get the result r4 = 4r.sin((A+B)/4).sin((B+C)/4).sin((C+A)/4)? Can anyone explain it to me, please? Thank you.
1)triangles AOF and AOD are congruent
ReplyDeletethe bisector of angle DAF is also the bisector of the central angle DOF,M midpoint of arc DF lies on this bisector
M lies on the bisector of inscribed angle AFD whose leg AF is tangent to the incercle
the bisectors of triangle ADF meet at M=O1
2)in triangle AOF
sin(A/2)=r/AO=r'/(AO-r)
r'=r(1-sin(A/2))
the same way
r''=r(1-sin(B/2);r'''=r(1-sin(C/2))
in triangle DEF,r is the circumradius and r4 is the inradius
r4=4rsin((A+B)/4)sin((B+C)/4)sin((C+A)/4)
after substitution and appropriate transformation
r'+r''+r'''+r4=2r
.-.
About the solution of problem 209 by Anonymous - December 31, 2009:
ReplyDeleteHow can I get the result
r4 = 4r.sin((A+B)/4).sin((B+C)/4).sin((C+A)/4)?
Can anyone explain it to me, please?
Thank you.