Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 700.
Monday, December 5, 2011
Problem 700: Equilateral Triangle, Circle, Circular Segment, Midpoint, Metric Relations
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ReplyDeleteFD cut AB and AO at G and H ( see picture)
Note that AD=AG=DG=6
OA= AG*2/√3 =12/√3
In triangle OHF , applying Pythagoras Theorem we get HF=3. √5
So x=HF-HD=3(√5-1)
Peter Tran
Extend ED to cut AC at X and circle at Y
ReplyDeleteEDXY ∥ AB
So DE = 6 ; and XY = x (by symmetry)
ED.DY = AD.DC
x(6 + x) = 36
x²+ 6x = 36
(x + 3)² = 45
x = 3(√5 - 1)