## Saturday, September 17, 2011

### #Geometry Problem 670: Triangle, 60 Degrees, Orthocenter, Incenter, Circumcenter, Congruence

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

Click the figure below to see the complete problem 670.

1. http://img835.imageshack.us/img835/1895/problem670.png
Per the result of problem 662
We have ( AHC)=(AIC)=(AOC)=120
And 5 points A, H, I, O and C are co cyclic ( see picture)
Since (B)=60 >> (BAH)=30
Since (AOC)=120 >> (OAC)=30
AI is and angle bisector of (BAC) >> AI is also an angle bisector of (HAO)
Arc HI=arc IO and IH=IO
Peter Tran

2. B = 60°, A + C = 120°.
Let D be the midpoint of BC
BH = 2 OD = 2R cos B = R = BO
∠HBA = 90° - A
∠OBC = ∠OCB = 90° - ∠BOC/2
= 90° - A = ∠HBA
Follows BI bisects ∠HBO and
∆BIH ≡ ∆BIO
∴HI = IO

3. Let D be midpoint of BC and let CH extended meet AB at E.

BE = a/2 = BD and < ABH = < OBD = 90-A
Hence tr.s BHE and BDO are congruent ASA and so BH = BO.

Now < HBI = < OBI = B/2 - (90-A)
Hence tr.s HBI and OBI are congruent SAS

Therefore HI = OI

Sumith Peiris
Moratuwa
Sri Lanka