Common tangents

Statement of the problem

Consider two circles, $C_1$ and $C_2$ with centers $O$ and $o$, as shown in the diagram above. Complete the rest of the diagram according the following prescription.

1. Let the external common tangent $AB$ intersect the extended line $Oo$ at point $P$.
2. Let the internal common tangent $CD$ intersect the line $Oo$ at point $Q$.
3. Let $M$ be the foot of the perpendicular dropped from $P$ onto $CD$.
4. Let $N$ be the foot of the perpendicular dropped from $Q$ onto $AB$.

Proposition 1: The line $MQ$ bisects that angle $OMo$, and the line $NQ$ bisects that angle $ONo$.

Proposition 2: The lengths of the line segment $PM$ and $QN$ are are independent of the distance between the centers of the circles.

I leave the proofs to you, the interested reader. To get you started, let us write $R$ and $r$ for the radii of the circles $C_1$ and $C_2$ respectively, and let $d$ be the distance between the circles' centers. The proof of Proposition 2 amounts to showing that \[ PM = \frac{2Rr}{R-r}, \quad QN = \frac{2Rr}{R+r}, \] and noting that these are independent of $d$. To prove Proposition 1, show that: \[ QO = \frac{Rd}{R+r}, \quad Qo = \frac{rd}{R+r}, \quad MO = \frac{R}{R-r} \sqrt{4Rr-d^2}, \quad Mo = \frac{r}{R-r} \sqrt{4Rr-d^2}. \] It follows that $QO/Qo = MO/Mo$, that is, the line $MQ$ subdivides the side $Oo$ of the triangle $OMo$ in proportion to the lengths of the sides $MO$ and $Mo$, which proves that it is the bisector of the angle $H$.

A note on the origin of this problem

The problem analyzed in this page was motivated by a question raised by <tlim1@cox.net> in the usenet newsgroup sci.math in 2003. Here is the relevant exchange:

%---- begin of usenet message -----------------------------------%

Newsgroup: sci.math
From: rouben@pc18.math.umbc.edu (Rouben Rostamian)
Subject: Re: A geometry question
Date: Sat, 4 Jan 2003 00:55:04 +0000 (UTC)

In article <gh7R9.114916$pe.4464622@news2.east.cox.net>,
TCL <tlim1@cox.net> wrote:
>In figure
>I believe l_4 bisects angle O_1BO_2.
>Is this true?
>If so, how to prove it?

Although it is said that a picture is worth a thousand words,
pictures are inconvenient in Usenet's text-only medium.
Therefore I will first translate your picture into words,
then I will outline a solution.

--- Statement of the problem ----------------------------------

Consider two circles with centers A and B and radii a and b.
Let d be the distance between the centers.  Assume d > a + b,
that is, the circles are outside each other.

Let a common tangent T1 to the two circles intersect the extended
line AB at a point D outside the segment AB.

Let another common tangent T2 to the two circles intersect the line
segment AB at a point E between A and B.

Let H be the foot of the perpendicular from D dropped onto T2.

Show that HE bisects the angle H of the triangle AHB.

--- Solution --------------------------------------------------

A straightforward computation shows that:

EA = ad/(a+b)
EB = bd/(a+b)
HA = [a/(a-b)] sqrt(4ab-d^2)
HB = [b/(a-b)] sqrt(4ab-d^2)

whence  EA/EB = HA/HB.

Thus the line HE subdivides the side AB of the triangle AHB in
proportion to the lengths of the sides HA and HB, proving that
it is the bisector of the angle H.

-- 
Rouben Rostamian <rostamian@umbc.edu>

%---- end of usenet message -----------------------------------%

This page was created on January 3, 2003 and was last updated in February 2020.