## Friday, May 18, 2012

### Problem 754: Equilateral Triangle, Center, Angle, 60 Degrees, Perimeter

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

This entry contributed by Ajit Athle.
Click the figure below to see the problem 754 details. 1. http://img100.imageshack.us/img100/2228/problem754.png

Let I is the incenter of triangle DBE ( see sketch)
∠DIE= ∠B+∠D/2+∠E/2= 60+ (180-60)/2= 120
Since ∠DIE supplement to ∠DOE , quadrilateral DIEO is cyclic
O is the intersecting point of angle bisector BI and circumcircle of DIE => O must be excenter of triangle DBE
Let excircle touch DE at T .
We have EQ=ET and DP=DT
And perimeter of DBE=BP+BQ= AC

2. Rotate triangle ODB 120(degree) clockwise about O,
it becomes triangle OD'B', where D' lies on BC and B'=C.

Now since OD=OD', and angle D'OE=120(degree)-angle DOE=60(degree),
thus triangle DOE is congruent to triangle D'OE.
Hence, DE=D'E.

Therefore, perimeter = BE+ED+DB = BE+ED'+D'B' = BC.

3. Excellent proof Jacob Ha!

1. Thank you very much!=)

4. http://www.mathematica.gr/forum/viewtopic.php?f=20&t=27343&p=133800

5. Problem 754
Let the point F on the side BC (F is between the points B, C), such that BD=FC.Then
Triangle DBO=triangleFCO (OB=OC,DB=FC, <DBO=<FCO=30).So OD=OF and <DOB=<COF.
But <COB=120=<DOF.Then <EOF=60=<EOD.So triangleDOE=triangleEOF and DE=EF.So
AC=BC=BE+EF+FC=BE+ED+DB.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

6. Let circle OBE cut AB at F.

Considerig cyclic quadrilateral BEOF, OF = OE and < FOD = 60. So Tr.s FOD and EOD are congruent SAS and so DE = DF.

Now Tr.s AOF and BOE are congruent ASA and hence AF = BE

So BD + DE + BE = BD + DF + AF = AB = AC

Sumith Peiris
Moratuwa
Sri Lanka

7. Solution 2

Let circle OBE cut BC at F

Note that OE = OF = a say and EF = sqrt3. a
Note also that DE = DF

Applying Ptolemy to cyclic quadrilateral BEOF and simplifying we get

Perimeter of Tr. BDE = OB. Sqrt3 = AC