Problem:
Triangle ABC has side lengths AB=4,BC=5, and CA=6. Points D and E are on ray AB with AB<AD<AE. The point Fξ =C is a point of intersection of the circumcircles of β³ACD and β³EBC satisfying DF=2 and EF=7. Then BE can be expressed as da+bcββ, where a,b,c, and d are positive integers such that a and d are relatively prime, and c is not divisible by the square of any prime. Find a+b+c+d.
Solution:
Because quadrilateral DFAC is cyclic, β DFC=β DAC=β BAC. Because quadrilateral EFBC is cyclic, β EFC=β EBC=180βββ ABC.
Hence β EFD=β EFCββ DFC=180βββ ABCββ BAC=β ACB. Applying the Law of Cosines to β³DEF and β³ABC gives
cosβ EFD=cosβ ACB=2β 6β 562+52β42β=43β
and
DE=72+22β2β 2β 7β 43ββ=42β
Let CF intersect line AB at G. Because β³ACGβΌβ³FDG and β³BCGβΌβ³FEG,
3=FDACβ=DGCGβ=GFGAβ and 75β=FEBCβ=EGCGβ=GFGBβ.
Therefore
521β=DGEGβ=DGED+DGβ and 521β=GBGAβ=GBGB+BAβ.
Solving for DG and GB yields GD=452ββ and BG=45β, so
