## Tuesday, July 10, 2012

### Problem 781: Triangle, Orthocenter, Circumcenter, Circle, Vertex, Altitude, Midpoint, Perpendicular

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

Click the figure below to see the complete problem 781.

1. A trigonometric Proof: (sketch)
Let∠DMB=θ
In ΔOBM:R/cosθ = R cosB/sin(90°–θ–C+A)
= R cosB/cos(C–A+θ),
cosB cos θ = cos(C-A)cosθ – sin(C–A)sinθ,
(*) tanθ = 2cosC. cosA/sin(C–A)
In ΔBDM: DM/cosA = R cosB/ sin(θ+90°–A)
= Rcos B/cos(A – θ)
In ΔBEM: ME/cosC = RcosB /sin(θ–(90°–C))
= - Rcos B /cos(C+θ)
It suffices to prove:
cosA cos(C+θ)+cosC.cos(A-θ)= 0 (or)
cosθ(2cos C.cos A)= sinθ.sin(C – A)
This follows from (*)
Hence DM = ME

2. Let N be the midpoint of AB. Then NM || AH and ON || CH so angle(ONM)=angle(C1HA)=B.
But ONDM is inscrided, then angle(ONM)=angle(ODM)=B.
The same is true for angle OEM. So DO=DE.

1. So DO=EO, therefore OM is the perpendicular bisector of DE and M is the midpoint of DE.

3. Anonymous

You got an excellent solution. !
attached is the sketch per your solution
http://img9.imageshack.us/img9/3192/problem781.png

Peter Tran