2018 AIME I Problems/Problem 15: Difference between revisions
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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 | |||
Revision as of 18:43, 9 March 2018
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
