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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 927.
FG = 2R sin80°tan10°PQ = 2R sin60°tan10°MN = 2R sin40°tan10°KL = 2R sin20°tan10°FG + PQ + MN + KL= 2R tan10° [sin20°+sin40°+sin60°+sin80°]= 2R tan10° [4 sin50°cos20°cos10°]= 8R cos20°cos40°cos80°= 8R [sin20°cos20°cos40°cos80°]/sin20°= 4R [sin40°cos40°cos80°]/sin20°= 2R [sin80°cos80°]/sin20°= R sin160°/sin20°= R
To Jacob: Thanks for your solution. Now, try an elementary geometry (Euclid's Elements.) solution.
http://img849.imageshack.us/img849/7236/vn8u.pngDefine points C’, D’,E’,F’, C1, D1,E1, S, T and U as per sketchDue to sysmetrical proprety, EFF’E’ is a isoceles trapezoid with its diagonals intersect at U on diameter AB.∠ (EUF)=1/2(Arc(EF)+Arc(E’F’))= 20°∠ (FUO)= ∠ (EUA)=80°So UE//OF and UF//OGSimilarly we have SC//TD//UE//OF and SD//TE//UF//OGFrom this proprety we have FG=OU, PQ=EE1=UT, MN=DD1=TS, KL=CC1=ASSo FG+PQ+MN+KL=OU+UT+TS+SA= radius R
The following facts could be easily found so i skip the proof:EF//DG//CH//AI & FG//EH//DI//CJ//ABSo FEPG,EDMQ,DCKN,CAOL are parallelogram.In addition, observe that∆FEP~∆PDM~∆MCK~∆KAO~∆FOG~∆POQ~∆MON~∆KOLNow the proof starts:1. OA = OF = RSo, ∆KAO = ∆FOG -> OK=FG2. EF=FG=OK (from 1)So, ∆FEP = ∆KOL -> FP=KL3.CK=CL-KL=OA-KL=OF-FP=OPSo, ∆MCK=∆POQ -> MK = PQ4.∠MOD=∠FOD=40=∠DOA=∠MDO, so DM=MOSo, ∆PDM=∆MON -> PM = MNSumming up the result of 1,2,3,4, FG + PQ + MN + KL = OK + FP + MK + PM = OK + KM + MP + PF =Rq.e.d.