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.

## Friday, May 18, 2012

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

Labels:
60 degrees,
center,
equilateral,
perimeter,
triangle

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http://img100.imageshack.us/img100/2228/problem754.png

ReplyDeleteLet 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

Rotate triangle ODB 120(degree) clockwise about O,

ReplyDeleteit 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.

Excellent proof Jacob Ha!

ReplyDeleteThank you very much!=)

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

ReplyDeleteProblem 754

ReplyDeleteLet 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