Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 778.
Monday, July 2, 2012
Problem 778: Triangle, Distance from the Orthocenter to a Vertex, Circle, Circumradius, Side, Square
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http://img213.imageshack.us/img213/5918/problem778.png
ReplyDeleteDraw lines per attached sketch
We have LC//BH and LB//CE
So BLCH is a parallelogram and
BH=LC
In right triangle ACL we have AC^2+CL^2=AL^2
Or BH^2+AC^2=4.R^2
http://img232.imageshack.us/img232/3764/p778resuelto.png
ReplyDelete(look the construction of the picture).
As AP is diameter, <PCA = 90° and <ABP=90°. Note that PC and HC are paralell to BH and BP respectively. Then BH = PC (because BPCH is paralelogram) and by pythagorean theorem we have that PC^2 + AC^2=4R^2. Hence BH^2+AC^2=4R^2
BH = 2R cos(ABC)...by Prob 777
ReplyDeleteAC = 2R sin(ABC)...by Sine rule
BH^2 + AC^2 = 4R^2
By using the result in problem 777, the statement would be reduce to AC = 2R sinABC
ReplyDelete,which is obvious under sine law.
Let AO meet the circle at X.
ReplyDelete< AXB = < B so < BCE = 90. - B = < CAX = CBX hence BX//HC and since BF//XC BXCF is a parellelogram and it thus follows that BH = XC
Now applying Pythagoras to right Tr. ACX the result follows
Sumith Peiris
Moratuwa
Sri Lanka