2018 AIME I Problems/Problem 15
Revision as of 18:43, 9 March 2018 by Specialbeing2017 (talk | contribs)
Basically Three cases are pretty similar WLOG ABCD ACDB ADBC As the diameter is 2 Let arcAB =2a arcBC=2b arcCD=2c arcDA=2d a+b+c+d=Pi By easy angle chasing phiA=a+c or b+d phiB=a+b or c+d phiC=a+d or b+c AC of case 1 = 2sin(arcABC/2)=2sin(a+b) BD of case 1= 2sin (arcBCD/2)=2sin(a+d) Sin(angleAC,BD of case 1)=sin(a+c) Area=AC*BD*sin(phiA)/2=2sin(phiA)sin(phiB)sin(phiC)=24/35 ANS=24+35=059
