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Two Circles, Four Tangents, Collinear Midpoints

Date: 12/20/98 at 05:19:04
From: hai trung ho
Subject: Geometry

Given two circles that do not touch, there are four distinct tangents 
common to both circles. Prove that the midpoints of the tangents are 

Date: 12/24/98 at 07:25:56
From: Doctor Floor
Subject: Re: Geometry

Hello Hai Trung Ho,

Thanks for your question.

To make the explanation easier I have made a picture for you:

We will repeatedly use the fact that two tangents to one circle from 
the same point have the same length, and of course we will use the 
symmetry in the line DA.

As a first consequence we can introduce the following names for 
segment lengths:

   p: = OT = TI = EN = HN
   q: = BS = SQ = LR = RK

We will show first that in fact two names were not necessary, because 
p = q.

To see this, realize that TL = TQ = OQ-p. And thus IL = TL-p = OQ-2p.

But also RI = RH = KH-q = OQ-q. And thus IL = RI-q = OQ-2q.

Since IL = OQ-2q = OQ-2p, apparently p = q.

But then the midpoint of HK is equal to the midpoint of NR, because 
from both the left and the righthand side the same length is taken. In 
the same way, the midpoint of EB is equal to the midpoint of NS. 

Since the two segments NR and NS both start from point N, the line 
connecting their midpoints must be parallel to RS and this line must be 
perpendicular to DA, since RS is.

Conclusion: the line through the midpoints of HK and EB is 
perpendicular to DA.

Because of the symmetry of the complete figure in DA, this conclusion 
implies that the line through the midpoints of IL and OQ is 
perpendicular to DA too, having the same point of intersection with 
DA. But then the four midpoints must be on one and the same line! And 
that proves your question.

Best regards,

- Doctor Floor, The Math Forum   
Associated Topics:
High School Conic Sections/Circles
High School Euclidean/Plane Geometry
High School Geometry

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