Geometry Problem

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

Click the figure below to see the complete problem 777.

## Sunday, July 1, 2012

### Problem 777: Triangle, Distance from the Orthocenter to a Vertex, Circle, Circumradius, Cosine

Labels:
altitude,
circumcenter,
circumradius,
cosine,
distance,
orthocenter,
triangle

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http://img839.imageshack.us/img839/7122/problem777.png

ReplyDeleteDraw lines per attached sketch

We have LC//BH and LB//CE

So BLCH is a parallelogram and

BH=LC=AL. cos(ALC)= 2.R.Cos(ABC)

HB = BD / sin(BHD)

ReplyDelete= AB*cos(ABC) / sin(ACB)...CDHF concyclic

= 2R*cos(ABC)...Sine rule

AC/sin(ABC) = 2R (R is the circumradius)

ReplyDeleteThe statement is then reduced to prove :

AC/sin(ABC) = BH/cos(ABC) <=> AC = BH*tanABC

which is the properties of the altitude

q.e.d.

Let M be the midpoint of AC.

ReplyDeleteMedian BM intersects Euler line OH at centroid G.

∆BGH ~ ∆MOG since OM ∥ BH

So BH/OM = BG/GM = 2 and

BH = 2OM = 2.OA.cos∠AOM = 2R cos∠ABC

[since∠AOM=(1/2)∠AOC=(1/2)(2∠ABC)=∠ABC]

Since < BHD = < C, BD = BH sin <C ...,(1)

ReplyDeleteAlso AB = 2R sin <C ....(2)

And BD = AB cos <B ...(3)

Eliminating AB and BD between these 3 equations

BH = 2R cos < B

Sumith Peiris

Moratuwa

Sri Lanka

Alternatively let AO meet the circle at X then prove that BHCX is a parellelogram (refer my proof of Problem 778)

ReplyDeleteThen BH = CX = 2R cos <B since < AXC = < B