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.