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
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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