Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 679.
Saturday, October 22, 2011
Problem 679: Triangle, Circle, Median, Proportion
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http://img580.imageshack.us/img580/7519/problem679.png
ReplyDeleteDraw RMS // DFE ( see picture)
Note that Area(ABM)=area(BMC) ( same altitudes and bases )
Area(ABM)= Area(BMC)=½* c*MN =1/2*a*MP
So a/c=MN/MP
Note that triangle RBS is isosceles and triangle MNR similar to tri. MPS
So MN/MP=MR/MS=FD/FE=x/y
So a/c=x/y
Peter Tran
let be r the radius,B1,B2 the angles at vertex B
ReplyDeleteusing the sine law
in DBF:sinB1/x=sinF/r
in EBF:sinB2/y=sinF/r
in ABM:sinB1/(0.5b)=sinA/BM
in CBM:sinB2/(0.5b)=sinC/BM
in ABC:sinA/a=sinC/c
hence a/c=x/y
Draw AG//DE and CH//DE, giving K and L the intersections with BM. r = radius circle.
ReplyDeletehttp://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/r
Because LC = AK, this gives ay/r = cx/r
=> ay = cx => a/c = x/y
QED.
Draw a line CPQ parallel to EFD intersecting BM at P and AB at Q
ReplyDeleteSince BDE and BQC are similar isosceles triangles, CP = ky and PQ = kx
Applying Menelaus to triangle ACQ
=> (BQ/BA)*(AM/MC)*(CP/PQ) = 1
=> (a/c)*(1)*(ky/kx) = 1
=> a/c = x/y
BM is the median
ReplyDeleteAM=MC & [ABM]=[CBM]
(AB)(BM)sin sin<DBF/x=sin<EBF/y
x/y=sin<DBF/sin<EBF=sin<ABM/sin<CBM=a/c