That problem with many little circles in a crescent.
David Karr
>The problem was:-
>----------------
>There are two circles internally tangent, with a doubly infinite sequence
>of small circles in the crescent between, each one tangent to the two
>original circles, and to its two neighbors in the sequence.
>
>Prove that the centres of the little circles all lie on yet another circle.
>----------------
>
>Alas, the statement is FALSE.
I thought the problem was to prove that the points of tangency
(between adjacent small circles) lay on yet another circle.
I don't recall any claim about the centers of the circles.
-- David A. Karr (ka...@cs.cornell.edu)
Bill Taylor
The posted solution was wrong; and I don't recall seeing a followup,
so here goes. Apologies if it's already appeared.
The problem was:-
----------------
There are two circles internally tangent, with a doubly infinite sequence
of small circles in the crescent between, each one tangent to the two
original circles, and to its two neighbors in the sequence.
Prove that the centres of the little circles all lie on yet another circle.
----------------
Alas, the statement is FALSE.
The proof given was to perform an inversion on the whole diagram about
some suitable circle so that the two original circles become parallel lines
and the little circles lined up between, with their centres clearly on
a straight line.
Unfortunately, though circles invert to (lines or) circles, their centres
do NOT invert to the centres!
I would say "go to the bottom of the class"; except that I maintained the
same mistake myself for some time.
The fact is, the centres of the little crescent-circles lie on an ellipse,
which is *never* a circle, (except in the degenerate limiting case).
I won't attempt an ascii-proof, but you can jot the diagram down simply
enough from the following.
Centre the first two circles on the positive x-axis with centres at A,B and
tangency point at O. Make it that the A circle is the smaller.
Draw a little crescent circle above right, with centre at C.
Let the A,B circles have radii r,R respectively, and the little one
radius p. Then AC = r + p
BC = R - p
So AC + BC = R + r = constant. Thus the locus of C is an ellipse with
foci at A & B.
[TA-DAH] <--- (ascii has no proof-end symbol, so I use this.)
----------
But now further. If you peruse your diagram, (all those who screwed it
up into the waste-basket take two demerit points!), you will note that O,
the point of tangency, plays no role in the proof! In fact, the theorem
works for any two circles, tangent or not.
And more, it can be extended to having the little circles touching one
or other of the original pair internally/externally, and one still gets a
conic section, though half the time it's a hyperbola! Neato.
And one can get a parabola, by making one of the originals a straight line!
So in fact the whole problem, while intriguing, has nothing to do with
inversion, but is virtually a mere illustration of the definitions of
conic sections by sums/differences-of-lengths. Which I often feel should
be "THE" geometrical definition of them anyway. It's a much simpler
definition for one thing, 2-dimensional rather than 3. Then one can also
prove their section-hood (and more) by that amusingly simple theorem I
mentioned in the "favorite short proofs" thread!
Fun.
-------------------------------------------------------------------------------
Bill Taylor w...@math.canterbury.ac.nz
-------------------------------------------------------------------------------
Every problem has at least one solution which is elegant, neat - and wrong.
-------------------------------------------------------------------------------
P.S. Was a sig-line ever so appropriate for its post!
Wei-Hwa Huang
>This little problem appeared in math or puzzles, I forget which.
>The posted solution was wrong; and I don't recall seeing a followup,
>so here goes. Apologies if it's already appeared.
>The problem was:-
>----------------
>There are two circles internally tangent, with a doubly infinite sequence
>of small circles in the crescent between, each one tangent to the two
>original circles, and to its two neighbors in the sequence.
>Prove that the centres of the little circles all lie on yet another circle.
>----------------
>Alas, the statement is FALSE.
>The proof given was to perform an inversion on the whole diagram about
>some suitable circle so that the two original circles become parallel lines
>and the little circles lined up between, with their centres clearly on
>a straight line.
>Unfortunately, though circles invert to (lines or) circles, their centres
>do NOT invert to the centres!
<shrug>
I guess I go to the back of the class.
Can I get some points back by pointing out the the tangent points of all the
little circles do indeed lie on a circle?
--
Wei-Hwa Huang, whu...@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
How many words can you find that are anagrams of themselves?
Hint: There are at least ten.
Ilias Kastanas
Bill Taylor <w...@math.canterbury.ac.nz> wrote:
>This little problem appeared in math or puzzles, I forget which.
>
>The posted solution was wrong; and I don't recall seeing a followup,
>so here goes. Apologies if it's already appeared.
>
>The problem was:-
>----------------
>There are two circles internally tangent, with a doubly infinite sequence
>of small circles in the crescent between, each one tangent to the two
>original circles, and to its two neighbors in the sequence.
>
>Prove that the centres of the little circles all lie on yet another circle.
>----------------
>
>Alas, the statement is FALSE.
>
>The proof given was to perform an inversion on the whole diagram about
>some suitable circle so that the two original circles become parallel lines
>and the little circles lined up between, with their centres clearly on
>a straight line.
>
>Unfortunately, though circles invert to (lines or) circles, their centres
>do NOT invert to the centres!
>
Well, Bill, I am one to uphold definitions that go back to Apollonius
of Perga! But I confess there is stiff competition from von Staudt. With
the benefit, admittedly, of several centuries, the "intersections of lines
corresponding, given two projective pencils" does take the (ultimate?)
step in unification.
OK, maybe it goes too far. I can still recall my professor discussing
the conic whose conjugation involution coincides with the orthogonality
involution (harmonic conjugates w.r.t. the circular points at infinity. In
real projective geometry all conics "look" the same, but some intersect the
line at infinity and some don't. What?? Enter complex projective geometry,
and they all do!) It has to be a hyperbola. With visible relish, he said
"Let us consider a hyperbola"... and drew an ellipse on the board.
Beholding your crescent from those lofty heights -- ellipse, circle,
what does it matter, all are one!
The circles' centers partake of metric crudeness ... they should be
content to sit on an ellipse, it is more than they deserve anyway. The way
I remember it, though, the problem involved the points where the circles
touch! Of course tangency has impeccable projective credentials; it is just
incidence, so pure! And sure enough, those points are on a circle.
Ilias