Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 679.
http://img580.imageshack.us/img580/7519/problem679.pngDraw RMS // DFE ( see picture)Note that Area(ABM)=area(BMC) ( same altitudes and bases )Area(ABM)= Area(BMC)=½* c*MN =1/2*a*MPSo a/c=MN/MPNote that triangle RBS is isosceles and triangle MNR similar to tri. MPSSo MN/MP=MR/MS=FD/FE=x/ySo a/c=x/yPeter Tran
let be r the radius,B1,B2 the angles at vertex Busing the sine lawin DBF:sinB1/x=sinF/rin EBF:sinB2/y=sinF/rin ABM:sinB1/(0.5b)=sinA/BMin CBM:sinB2/(0.5b)=sinC/BMin ABC:sinA/a=sinC/chence a/c=x/y
Draw AG//DE and CH//DE, giving K and L the intersections with BM. r = radius circle.http://s73.photobucket.com/albums/i207/HenkieR/PuzzleSolutions/GoGeometryProblem679solution_rlg.jpg∆MCL = ∆MAK (opposite angles, Z-angles and AM=CM)so LC = AK.∆BEF ~ ∆BCL (aa), so y/r = LC/a => LC = ay/r∆BDF ~ ∆BAK (aa), so x/r = AK/c => AK = cx/rBecause LC = AK, this gives ay/r = cx/r=> ay = cx => a/c = x/yQED.