Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 854.
Thursday, February 7, 2013
Problem 854: Triangle, Parallel lines, Parallelogram, Areas, Similarity, Concurrent lines
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(FBO)/S1=FB/FD=OE/OD=EM/MB=S2/(OBM).So, (FBO).((OBM)=S1.S2
ReplyDelete(S5/2)^2= S1.S2 ,S5=2sqrt(S1.S2) (1)
(OEC)/S2=CE/EM=HO/OM=HG/GC=S3/(OGC).So (OEC).(OGC)=S2.S3
(S6/2)^2=S2.S3 , S6=2sqrt(S2.S3) (2)
(AHO)/S3=AH/HG=FO/OG=FD/DA=S1/(ADO).So ,(AHO).(ADO)=S1.S3
(S4/2)^2=S1.S3 , S4=2sqrt(S1.S3) (3)
(1)+(2)+(3) S5+S4+S6=2sqrt(S1.S2) +2sqrt(S2.S3) +2sqrt(S1.S3)
Plan : http://img16.imageshack.us/img16/6624/p854triangleparallelogr.gif
Let S be the area of triangle ABC.
ReplyDeleteS1 = S*(FD/AB)^2
S3 = S*(DA/AB)^2
=> 2sqrt(S1S3) = 2S*(FD/AB)*(DA/AB) = 2S*(OD/AC)*(AD/AB) = S4
Symmetrically,
2sqrt(S2S3)= S6
2sqrt(S1S2)= S5
Q.E.D.