Geometry Problem

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

Click the figure below to see the complete problem 781.

## Tuesday, July 10, 2012

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

Labels:
altitude,
circle,
circumcenter,
circumcircle,
midpoint,
orthocenter,
perpendicular,
triangle,
vertex

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A trigonometric Proof: (sketch)

ReplyDeleteLet∠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

Let N be the midpoint of AB. Then NM || AH and ON || CH so angle(ONM)=angle(C1HA)=B.

ReplyDeleteBut ONDM is inscrided, then angle(ONM)=angle(ODM)=B.

The same is true for angle OEM. So DO=DE.

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

DeleteAnonymous

ReplyDeleteYou got an excellent solution. !

attached is the sketch per your solution

http://img9.imageshack.us/img9/3192/problem781.png

Peter Tran