## Thursday, January 14, 2016

### Geometry Problem 1179: Triangle, Centroid, Transversal, Sum of Ratios, Sides

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1179.

1. http://s9.postimg.org/71ehki32n/pro_1179.png

Draw median BM and lines AL//DE and CN // DE
Note that triangles ALM and CNM are congruence… ( case ASA)
We have a1/a2= GL/GB=(GM-ML)/GB= ½ - ML/ GB …. (1)
C1/c2= GN/GB= ( GM+MN)/GB = ½ + MN/GB …. (2)
Add ( 1) to (2) and note that ML=MN we get the result.

2. MANOLOUDIS APOSTOLIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

BG intersecting AC M is AM=MC and BG=2.MG ,si DE intersecting AC K .
Apply the theorem Menelaus tri ABM and tri BMC intersecting by KDGE.
Therefore AK/KM.MG/GB.BD/DA=1 or DA/DB=AK/2.KM and MK/KC.EC/EB.BG/GM=1 or
EC/EB=KC/2.MK. Therefore DA/DB+EC/EB=AK/2.MK+KC/2.MK=(AK+KC)/2.MK=(MK-AM+KM+MC)/2.MK=2.MK/2.MK=1.

3. Vijaya Prasad Nalluri (Pravin) Rajahmundry, INDIA.

Let M be the midpoint of AC. Let K, L be the points on the ray BM such that AK//DE//CL.
Note KL and AC bisect one another at M since AK//LC (ALCK is a parallelogram and so KM = ML)
Hence a1/a2 + c1/c2 = KG/BG + LG/BG = (KG +LG)/BG
= (KG + LM +MK + KG) /BG = (2 KG + 2KM)/BG
= 2GM/BG = BG/BG = 1

4. Extend DE to meet AC at P and M be the center of AC
let PA=x and AP=PC=z
Apply menelaus to ABM
=>(PM/AM)(a1/a2)(2/1)=1
=> (x+z/x)=a2/2a1
=> (x+z) = x.a2/2a1 ---------(1)
and x/z= a2-2a1/2a1-----(2)

Apply menelaus to ABC
=> (x+2z/x)(a1/a2)(c2/c1)=1
=> (x+2z)(a1/x.a2)=c1/c2
=> (x+2z)(1/x+z)=c1/c2 (from (1))
=> 1+1/(1+x/z)=c1/c2
=> 1+(a2-2a1/a2)=2c1/c2 (from (2))
=> a1/a2+c1/c2=1

5. Does the converse hold good?
Given D on BA and E on CB such that
AD = a1, DB = a2, CE = c1, EB = c2
satisfying a1/a2 + c1/c2 = 1 and
M is the midpoint of AC,
is it true that the segments AC and BM
intersect at the centroid ?

1. Yes, the converse does hold .
for example Given D on BA and E on CB such that
AD = a1, DB = a2, CE = c1, EB = c2
satisfying a1/a2 + c1/c2 = 1 then DE will pass through the centroid G of triangle ABC.

see sketch below :
https://photos.app.goo.gl/60Z3JPEl6dCV325Q2

Let G is the centroid of triangle ABC.
DG meet BC at E' . Denote CE'= c1' and BE'=c2'
per the result of problem 1179 we have a1/a2+c1'/c2'= 1
compare with original expression we get c1/c2= c1'/c2'
in segment BC there exist only one point E for each ratio EC/EB so E must coincide to E'.