## Wednesday, June 26, 2013

### Problem 892: Triangle, 60 degrees, Orthocenter, Circumcenter

Geometry Problem. GeoGebra, HTML5 Animation for iPad and Nexus.
Post your solution in the comments box below.

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

Click the figure below to see the dynamic geometry demonstration of problem 892.

1. http://img15.imageshack.us/img15/6545/fyi.png
let H’ and O’ are point of symmetry of H and O over AC.
H’ and O’ will be on circle O ….( property of orthocenter and angle AOC= 120)
Due to symmetric property, O and H will be on circle O’, radius=OO’ ( see sketch)
We have Angle(AOO’)=angle(AO’O)=60
And angle(AHO)= ½*angle(AO’O)=30

2. Extend EC, EC meet the circle at anoter point, T
HCF=EBF
HCF=TBE
TBE=EBH && BE Orthogonal to HT => TE=EH
=> AH=AT
ATH=60 => ATH Equilateral triangle
let M midPoint of AT
build HM, HM median and Altitude to AT=> O on HT =>
OH an angle bisector => OHA=30

3. Problem 892
Is <AOC=120=<AHC (<AHC=180-<ABC=180-60=120). Therefore AOHC is concyclic. So
<AHO=<ACB=(180-120)/2=30.