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 Subject: "circle of similitude" Previous Topic | Next Topic
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alexb
Charter Member
1606 posts
Jul-31-05, 07:54 PM (EST)

"circle of similitude"

 I've been looking into the properties of the circle of similitude of two given circles. It's going to be useful in a proof of Malfatti's problem. My knowledge comes mostly from Coolidge's book cited athttps://www.cut-the-knot.org/Curriculum/Geometry/CircleOfSimilitude1.shtmlAt some point, the book says, "It follows at once from the definition ..." What follows is rephrased as Statement 4 in the above page. I've been lookng back and forth for a few hours now, but do not see anything immediate. Does any body see what I am missing.Thank you,Alex

mr_homm
Member since May-22-05
Aug-01-05, 07:33 AM (EST)

1. "RE: circle of similitude"
In response to message #0

 Alex,I've been looking at this on and off, and I finally think I see what to do. Referring to the diagram in your applet, since line TL1 is tangent to circle 1 and L1K1 is secant to circle 1, it follows that the angle K1O1L1 is twice angle TL1L2. Let's call these angles x and 2x. Therefore |K1L1| = 2*R1*sin(x). However, the perpendicular distance h of T from line L1L2 is h = |TL1|*sin(x). Similarly, there are angles y and 2y at circle 2, and |K2L2| = 2*R2*sin(y), while h = |TL2|*sin(y).Now since |TL1|/|TL2| = R1/R2, we get the result that |K1L1| = 2*R1/|TL1|*h = 2*R2/|TL2|*h = |K2L2|. This is obviously reversible, so you get the converse too. I got this idea after playing with your applet for a long time and noticing that the distance from T to L1L2 seemed visually to vary in proportion to the lengths of the secants. From there, it was just a matter of trying to build a path of angle and line relationships between the secants and the perpendicular from T to L1L2.I must say, though, that this amount of fiddling is not my idea of "follows immediately from the definition." Also, now that I think of it, you could avoid reference to trig functions by noting that the triangle L1O1B1 is similar to TL1F, where B1 is the bisector of L1K1 and F is the foot of T on L1L2. Then you have a more "old school" type of proof.--Stuart Anderson

alexb
Charter Member
1606 posts
Aug-01-05, 05:40 PM (EST)

2. "RE: circle of similitude"
In response to message #1

 Thank you. This is simple enough.

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