Difference between revisions of "2018 AIME I Problems/Problem 15"
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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
