## Thursday, November 29, 2012

### Problem 828: Quadrilateral, Midpoint, Opposite sides, Diagonals, Concurrent lines, Triangle, Bimedian

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 828.

1. Connect EF,FG, GH and EH
We have FG//BD//EH and FG=1/2. BD=EH
So EFGH is a parallelogram so EG and FH intersect at midpoint of diagonal.
Connect MF,FN,NH and MH
We have FN//CD//MH and FN=1/2.CD=MH
So MFNH is a parallelogram so diagonals MN and FH intersect at midpoint of diagonal.
So EG, FH and MN are concurrent at the midpoint O

2. [Vector Method]
Let Vector(BA) = u ; Vector(BC) = v ; Vector (BD) = ru + sv , where r,s are scalars.
Hence
BE = (1/2)u ;
BF = (1/2)v ;
BH = (1/2)(BA + BD) = (1/2)[(r+1)u + sv]
BG = (1/2)(BC + BD) = (1/2)[ru + (s+1)v]
BM = (1/2)(u + v)
BN = (1/2)(ru + sv)

Now The mid point of MN = (1/2)*(BM + BN) = (1/2)*(1/2)*[(r+1)u + (s+1)v]
This can be rewritten into (1/2)*{(1/2)*u + (1/2)[ru + (s+1)v]} = (1/2)*(BE+BG)
i.e. the mid point of EG.
Symmetrically, it would also be the mid point of FH.

Q.E.D.

3. FM es base media en tr ABC así que FM // AB.
NH es base media en tr ADB así que HN // AB.
Finalmente GM // HN.
Análogamente FN // MH.
Entonces MFNH es paralelógramo y O es el pto medio de una diagonal, luego O pertenece a MN y OM=ON.

5. Let A,B,C,D are points of mass 1.

Since E,F,G,H,M,N are four mid-points, we have
E = 1/2(A+B)
F = 1/2(B+C)
G = 1/2(C+D)
H = 1/2(D+A)
M = 1/2(A+C)
N = 1/2(B+D)

Mid-point of EG = 1/2(E+G) = 1/4(A+B+C+D)
Mid-point of FH = 1/2(F+H) = 1/4(A+B+C+D)
Mid-point of MN = 1/2(M+N) = 1/4(A+B+C+D)

Let O=1/4(A+B+C+D)
Then EG, FH, MN are concurrent at their common mid-point O.