Geometry Problem. Post your solution in the comments box below.

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

Click the figure below to see the complete problem 879.

## Thursday, May 23, 2013

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

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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..........

ReplyDeleteDraw 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°

DeleteEC=X

ReplyDeleteDC=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

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

ReplyDeleteAngle 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

http://img208.imageshack.us/img208/6976/problem897.png

ReplyDeleteDraw GFH //BA

Since AF/AC=1/4 so AH=BG=GE=1/4.AD

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

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.

ReplyDeleteHence CDFE is cyclic and it follows that x = < FCD = 45.

Sumith Peiris

Moratuwa

Sri Lanka

Problem 879

ReplyDeleteIf 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