## Thursday, May 23, 2013

### Problem 879: Square, Midpoint, Diagonal, Ratio 3:1, Angle Measure

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

Click the figure below to see the complete problem 879.

1. point F is 1/4 of the diagonal AC, drop a vertical from E and a horizontal from F gives us a base of 1 and a height of 3 for the triangle giving us an atan of 1/3 for half of the angle E. From here we find the tanE = 36.8698976..........

1. Draw EG//AB meeting AC in G, let /_FEG=α & /_GED=β, x = α + β . Further, tan(α)=1/3 and tan(β)=1/2 and so tan(x)=tan(α+β)=[tan(α)+tan(β)]/[1-tan(α)*tan(β)]=[1/3 + 1/2] = [1-1/6]=1. Hence, x=45°

2. EC=X
DC=2X
build BD
AC=sqrt(5)x
AF=y
FC=3y
4y=sqrt(5)x
y=sqrt(5)x/4=AF
BD cut AC at Q
QF=sqrt(5)x/2-AF=sqrt(5)x/2-sqrt(5)x/4=sqrt(5)x/4
FQ=AF
DQ=sqrt(5)x/2
DC/DQ=EC/FQ=2x/0.5sqrt(5)x=x/0.25sqrt(5)x=
=4/sqrt(5)
FQD~ECD
DFQ=DEC
F point on circle of ECD
[can to prove it by "Reductio ad absurdum"]
FCD=FED=45

3. http://img829.imageshack.us/img829/3494/22778697.png

Angle EKF=angle KED=90+45=135 degrees.If,AB=a ,is, FK=AK/2=a.sqrt(2)/4,EK=a/2 ,KD=a.sqrt(2)
FK/EK=(sqrt(2))/2 (1) ,EK/KD=(sqrt(2))/2 (2)
From (1),(2) , FK/EK= EK/KD so, the triangles EKF,EKD is similar ,therefore, angle FEK= angle EDK=ω, angle KED= angle KFE=y. But ω+y=45 ,so, angle x=45 degrees

4. http://img208.imageshack.us/img208/6976/problem897.png
Draw GFH //BA
Triangle DHF congruence to FGE (case SAS)
So FD=FE and angle(GFE)+ angle (HFD)= 90 => angle (EFA)=90
Triangle EFD is a right isosceles triangle => angle (FED)=45

5. From congruent triangles we see that FB = FE = FD so F is the centre of the circumcircle of Tr. BED. So < EFD = 2 < EBD = 90.
Hence CDFE is cyclic and it follows that x = < FCD = 45.

Sumith Peiris
Moratuwa
Sri Lanka

6. Problem 879
If the BD intersects AC at O then CO=2AF=2FO or AF=FO.But OF/OD=1/2=EC/CD so triangle ECD is similar with triangle FOD. Then <CDE=<ODF but <CDE+<EDB=45 or
<ODB+<EDO=45 or <EDF=45=<ECF so ECDF is cyclic. Therefore <EFD=90 and <FED=45.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE