Two Circles, Four Tangents, Collinear MidpointsDate: 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 collinear. 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 http://mathforum.org/dr.math/ |
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