Friday, December 28, 2012

Problem 840: Parallelogram, Perpendicular, Diagonal, Metric Relations, Similarity

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

Click the figure below to see the complete problem 840.

1. build DM and BH altitudes TO AC [CMD=DHB=90]
DMC~AFC =>CM/FC=CD/AC
AC*CM=FC*CD
BHC~AEC=>BC/AC=HC/EC
EC*BC=HC*AC
AMD~=CHB =>CH=AM
EC*BC+CD*FC=AC*CM+HC*AC=AC(CM+HC)=AC(CM+AM)=AC*(AC)=AC^2
Q.E.D

2. By similar triangles ABE and ADF, we have
BE/AB = DF/AD <-> BE/CD = DF/BC <-> BE*BC=CD*DF

AC^2
=BC^2 + CD^2 + 2BC*AB*cos∠ABE
=BC^2 + CD^2 + 2BC*BE
=BC^2 + CD^2 +BC*BE + CD*DF
=BC*(BC+BE) + CD*(CD+DF)
=BC*CE + CF*CD
q.e.d.

3. Let G be a point such that EAGC is rectangle.
Let M be a point on AC such that DM⊥AC.

C,G,D,M concyclic ⇒ CE×BC = AD×AG = AM×AC
A,F,D,M concyclic ⇒ CF×CD = CM×AC

∴ CE×BC + CF×CD = AC×(AM+CM) = AC^2

4. Confining to ΔABC, in the usual notation,
b = a cos C + c cos A = a. CE/b + c. AF/b
AC² = b² = a.CE + c. CF = CB. CE + CD. CF

5. Prof Radu Ion,Sc.Gim.Bozioru,BuzauDecember 30, 2012 at 2:22 AM

BC^2=AF^2+FD^2=AF^2+(CF-DC)^2;DC^2=AE^2+EB^2=AE^2+(CE-BC)^2;
AC^2=AE^2+EC^2;AC^2=AF^2+FC^2

BC^2=AF^2+CF^2-2CF.DC+DC^2
DC^2=AE^2+CE^2-2CE.BC+BC^2
--------------------------------------------------
2CF.DC+2CE.BC=(AF^2+FC^2)+(AE^2+EC^2)=2AC^2

6. Solution to problem 840 by Michael Tsourakakis for Greece
ECFA,is ,cyclic quadrilateral.So EC.AF+AE.CF=AC.EF (1)More,angle AEF=angle ACF and
angle AFE=angle ECA=angle CAD.So,The triangles AEF, ACD are similar.Therefore
AC/EF=BC/AF=CD/AE.So,AE=CD.EF/AC and AF=BC.EF/AC
from (1) we,CE.BC.EF/AC+DC.EF.CF/AC=AC ,EF(CE.BC+DC.CF)=EF.AC.AC or
,CE.BC+DC.CF=AC^2