## Monday, June 16, 2008

### Elearn Geometry Problem 121

See complete Problem 121
Triangle, Similarity and Area. Level: High School, SAT Prep, College geometry

1. Consider Tr. BMC & AED as the transversal. We've
by Menelaus CD/DB*BE/EM*AM/AC=1 or BE/EM= 4/1 while BG/BM=2/3 or BE - BG=(4/5-2/3)BM or GE = (2/15)BM ----(1)
Now consider Tr. BNC & AFD as the transversal.
BA/AN*NF/FC*CD/BD=1 or 2/1*NF/FC*1/2=1 or NF/FC =1 or CF/CN=1/2 and CG/CN=2/3. Hence, CG-CF=(2/3-1/2)CN or GF==CN/6. Thus, GF/CF = (NC/6)/(CN/2)=1/3-------(2)
Now, Tr. GMC/Tr.BMC = 1/3 since GM/BM=1/3 and Tr. BMC=(1/2)Tr. ABC. Thus, Tr. GMC=(1/6)Tr. ABC
While Tr. GEF =(1/4)Tr.GEC becuase of (2)
=(1/4)(2/5)Tr. GMC becuase of (1)
=(1/4)(2/5)(1/6)Tr.ABC
or S1 =(1/60)S -----(3)
Ajit: ajitathle@gmail.com

2. we know that BE/EM=4/1,BG/BM=2/3
As BM=BE+EM,therefore BG/(BE+EM)=2/3
3BG=2BE+2EM
3BG=2EM(BE/EM+1)
3BG=10EM(BE/EM=4/1)
We also have GE+EM=BM-BG
EM(GE/EM+1)=BM-BG
((3/10)BG)(GE/EM+1)=BM-BG
GE/EM+1=(10/3)((BM-BG)/BG)
GE/EM=(10/3)(BM/BG-1)-1
GE/EM=(10/3)(3/2-1)-1
GE/EM=5/3-1
GE/EM=2/3

3. as I proved above that GE/EM=2/3,Now
in Tr.GMC Line DFA is transversal
therefore (CA/MA)(EM/GE)(GF/FC)=1
(2/1)(3/2)(GF/FC)=1
(GF/FC)=1/3

4. The following solution to problem 121 doesn’t use Menelaus’ theorem. As G is the center of gravity, then BG/GM = BD/DC = 2, so GD and AC are parallel. Besides, BD/BC = 2/3.
Since AME and DGE are similar and also BDG and BCM are similar, then
GE/EM = GD/AM = GD/MC = 2/3.
As GDF and CAF are similar, then
GF/FC = GD/AC = GD/(2MC) = (1/2)(2/3)=1/3.
If h is the altitude of ABC relative to the base b = AC, then CGM has base b/2 and altitude h/3, so S(CGM) = S/6.
Triangles CGM and FGM have same altitude relative to vertex M and, as GF = CG/4, then
S(FGM) = S(CGM)/4 = S/24.
Triangles EFG and FGM have the same base GF and GE/GM = 2/5, so
S1 = S(EFG) = (2/5)S(FGM) = (2/5)(S/24) = S/60.