## Sunday, July 1, 2012

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

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

Click the figure below to see the complete problem 777.

#### 6 comments:

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

Draw 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)

2. HB = BD / sin(BHD)
= AB*cos(ABC) / sin(ACB)...CDHF concyclic
= 2R*cos(ABC)...Sine rule

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

4. Let M be the midpoint of AC.
Median 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]

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

Also 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

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

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