Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 811.
Friday, October 5, 2012
Problem 811: Trisecting a Line Segment AB with four Circles and one Line, Radius, Center, Diameter, Chord
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Consider ∆DAE and ∆EAF.
ReplyDeleteThey are both isosceles and ∠DAE=∠EAF.
Therefore, ∆DAE ~ ∆EAF.
So AD / AE = AE / AF
But AD = CB + AB = 3*AB = 3*AE
Thus
AB / AF = AE / AF = AD / AE = 3
Hence, AF = AB / 3.
Triangles AEF and ADE are isosceles.
ReplyDeleteNote that ∠EAF=∠EFA=∠EAD=∠AED
So triangle AED similar to ADE ( case AA)
And AF/AE=AE/AD=1/3 => AF=1/3.AE=1/3.AB
Consider the inversion transform T = I(A, AB^2).
ReplyDeleteT(Circle A) = Circle A
T(Line CAB) = Line CAB
Now consider circle E.
T(circle E) would pass through the intersection points of circle A and circle E.
And since T(circle E) should be a line that does not pass through A,
so T(circle E) = the common chord of circle A and circle E.
Since the common chord of circle A and circle E
is exactly the perpendicular bisector of AE,
which should pass through point D.
T(circle E ∩ line CAB) = T(circle E) ∩ line CAB
Thus, T(F) = D.
By the definition of inversion,
AF * AD = AB^2
But AD = 3*AB,
hence, AF = AB / 3.