## Friday, October 5, 2012

### Problem 811: Trisecting a Line Segment AB with four Circles and one Line, Radius, Center, Diameter, Chord

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 811. 1. Consider ∆DAE and ∆EAF.
They 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.

2. Triangles AEF and ADE are isosceles.
So triangle AED similar to ADE ( case AA)

3. Consider the inversion transform T = I(A, AB^2).

T(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,