Proposition: Consider the circles $C_1$ and $C_2$ with centers $O$ and $o$, and their external tangents $AB$ and $CD$. Then the chords $AP$ and $QD$ cut by the line $AD$ are of equal lengths.
Proof: See the correspondence below.
From: "Rouben Rostamian" <rostamian@umbc.edu>
Date: Mon, 9 Aug 2004 14:43:32 -0400
To: "Poeppe Christoph" <Poeppe@spektrum.com>
Subject: Re: common tangents to two circles
Dear Dr. Poeppe,
I don't recall having seen the theorem that you have stated but I
was able to construct a simple proof for it.
The key to the proof is a theorem form elementary geometry known as
the "power of a point with respect to a circle". Here is a reminder
of what that theorem says:
Let A be a point outside the circle C. From A draw two rays.
One touches (i.e., is tangent to) the circle at T and the other
cuts the circle at points P and Q. Then AP . AQ = AT^2.
To prove the theorem that your reader has stated, let C1 be the
circle with points A and C on it, and let C2 be the other circle.
Let P and Q be the intersections of AD with C1 and C2, respectively.
Our purpose is to show that AP = QD.
Applying the theorem of power to the point A and circle C2 we get:
AQ . AD = AB^2.
Applying the theorem of power to the point D and circle C1 we get:
DP . DA = DC^2.
Since AB and DC are equal, we conclude AQ . AD = DP . DA.
Additionally, since AD and DA are the same segment, then it follows
that AQ = DP. Take away the common segment PQ from the two sides
of this equality to arrive at AP = DQ, as asserted.
With best regards,
Rouben Rostamian
--
Rouben Rostamian phone: 410-455-2412
Department of Mathematics and Statistics email: rostamian@umbc.edu
University of Maryland, Baltimore County fax: 410-455-1066
Baltimore, MD 21250, USA www: http://mathstat.umbc.edu
--- original message from "Poeppe Christoph" <Poeppe@spektrum.com>
Subject: common tangents to two circles
Date: Mon, 9 Aug 2004 12:16:47 +0200
From: "Poeppe Christoph" <Poeppe@spektrum.com>
To: <rouben@pc18.math.umbc.edu>
Dear Prof. Rostamian,
May I ask you for a few minutes of your attention?
A reader of our magazine (we are the German language translation of
"Scientific American") apparently found a theorem in Euclidean
geometry without being able to prove it. I confirmed it by brute
force: expressing everything in coordinates and verifying the
statement algebraically using Mathematica. But I didn't find any "lege
artis" proof. Your contribution to the sci.math newsgroup was the
closest one to this problem I could find on the Web.
The statement runs as follows.
Take two non-intersecting circles of different sizes and draw their
common outer tangents. They intersect at some point M on the common
symmetry axis of the whole figure and meet the two circles in points A
and B, and C and D respectively. (Primes refer to mirror images with
respect to the symmetry axis, so A and C are on one circle, and B and
D on the other.) Then the straight line AD (or CB) cuts chords of
equal length from both circles.
Did you ever see this kind of theorem? Or do you know about a source
to look into? Or does any idea about a proof come to your mind?
Your help would be very much appreciated.
Yours sincerely
Dr. Christoph Pöppe
Redaktion
Spektrum der Wissenschaft
Postfach 104840, D-69038 Heidelberg
Slevogtstr. 3-5, D-69126 Heidelberg
Tel (06221) 9126-719, fax -729
This page was created on August 9, 2004 and was last updated in February 2020.