Proposed Problem

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

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Complete Problem 434

Level: High School, SAT Prep, College geometry

## Thursday, April 15, 2010

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

Labels:
proportions,
quadrilateral,
ratio,
similarity,
transversal,
triangle

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BE/AE = CG/GD =>

ReplyDeleteBE/CG = AE/GD =>

BC//EG//AD =>

BE/AE = FO/HO = k

To c.t.e.o:

ReplyDeleteBC is not necessary parallel to EG and AD.

draw FP//EG (P on AB), CJM//EG (J on FH, M on AB)

ReplyDeleteHQ//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

--------------------------------------------

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.(*)

ReplyDeleteIn 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 (*).

BE/AE=CG/DG => EG//AD => FO/HO=BE/AE=k.

Similarly EO/GO=q.

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

using barycenter

ReplyDelete(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

.-.

It's very simple!

ReplyDeleteThe quadrilateral ABCD can be transformed a parallelogram ABCD with other conditions unchanged.

Denote Vec(AB) =Vector AB

ReplyDelete1. 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