## Thursday, April 15, 2010

### Problem 434: Quadrilateral, Transversal, Ratio, Measurement, Similarity

Proposed Problem
Click the figure below to see the complete problem 434 about Quadrilateral, Transversal, Ratio, Measurement, Similarity.

Complete Problem 434

Level: High School, SAT Prep, College geometry

1. BE/AE = CG/GD =>
BE/CG = AE/GD =>

BE/AE = FO/HO = k

2. To c.t.e.o:
BC is not necessary parallel to EG and AD.

3. draw FP//EG (P on AB), CJM//EG (J on FH, M on AB)
HQ//EG ( Q on AB), DKN//EG (K on FH, N on AB)

have to prove
OF/OH = CG/GD or EP/EQ = CG/GD ??? (1)

EP/EM = OF/OJ (PF//MJ//EG) (2)
EQ/EN = OH/OK (QH//NK//EG) (3)

from (2) & (3) (divide side by side)

(EP/EM)/(EQ/EN) = (OF/OJ)/(OH/OK)
=>
(EP/EQ)∙(EN/EM) = (OF/OH)∙(OK/OJ)
but
EN/EM = OK/OJ ( MJ//EG//NK)
=>
EP/EQ = OF/OH (1) is proved
--------------------------------------------

4. For any fixed point A, let O be a point on AB such that AO/BO is fixed. If B undergoes transformation, than O undergoes another transformation parallel to that of B with vector depends only on B's tranformation and AO:OB.(*)

In this problem, we fix A,E,B,q and let C,G,D undergo a transformation so that C moves to B, then F,O,H also undergo a transformation by (*).
Similarly EO/GO=q.

remarks: actually FO/OH = -k and EO/OG = -q.

5. using barycenter
(a+b+c+d)O=aA+bB+cC+dD=(a+b)E+(c+d)G
=(b+c)F+(a+d)H
let be a=1
aEA-bEB=0 then b=1/k
bFB-cFC=0 then c=qb=q/k
aHA-dHD=0 then d=q
(1+q)/kOF-(1+q)OH=0 then FO/OH=k
(k+1)kOE-(q/k+q)OG=0 then OE/OG=q
.-.

6. It's very simple!
The quadrilateral ABCD can be transformed a parallelogram ABCD with other conditions unchanged.

7. Denote Vec(AB) =Vector AB
1. Let O is a point on FH such that FO/OH=k and O’ is a point on EG such that EO’/O’G=q . We will prove that O coincide with O’ .

2. Let I is any point in the plane and E is a point on segment AB with Vec(BE)/Vec(EA) =k . Basic vector manipulation give us the formula Vec(IE)=1/(1+k) *Vec(IB) + k/(1+k) * Vec(IA)

3. Apply this formula for point G on segment CD, point F on segment BC and point H on segment AD.
Vec(IG)=1/(1+k) *Vec(IC) + k/(1+k) * Vec(ID)
Vec(IF)=1/(1+q) *Vec(IB) + q/(1+q) * Vec(IC)
Vec(IH)=1/(1+q) *Vec(IA) + q/(1+q) * Vec(ID)

4 Let m=(1+k)(1+q)
Apply formula in step 2 for point O of segment FH
Vec(IO)=1/(1+k)* Vec(IF) + k/(1+k) *Vec(IH)
Replace value of Vec(IF) and Vec(IH) in above equation we get.
Vec(IO)=1/m*Vec(IB) +q/m* Vec(IC) +k/m* Vec(IA) +k*q/m *Vec(ID)

5 Apply formula in step 2 for point O’ of segment EG
Vec(IO’)=1/(1+q)* Vec(IE) + q/(1+q) *Vec(IG)
Replace value of Vec(IE) and Vec(IG) in above equation we get
Vec(IO’)=1/m*Vec(IB) + k/m* Vec(IA) +q/m* Vec(IC) + k*q/m *Vec(ID)

6 From steps 5 and 6 we have Vec(IO=Vec(IO’) . This prove that O coincide with O’

Peter Tran
vstran@yahoo.com