Triangle puzzler with kissing circles
Jim Ferry
tangent to one of the triangles' sides, and which are mutually
externally tangent to each other and the triangle's incircle. We will
call this an "external kissing" configuration for the triangle.
In other words, these circles are the black circles in the "Kissing
circles" figure at
http://en.wikipedia.org/wiki/Descartes_theorem,
the triangle's incircle is the tiny red circle, and the triangle
itself must be drawn in without piercing any circles.
Such circles are not always possible. For example, consider an
isosceles triangle with apex angle alpha. When alpha = pi/3, (i.e.,
the triangle is equilateral), such circles clearly exist.
Problem 1 (easy): what is radius of these circles (as a multiple of
the triangle's inradius r)?
As we increase alpha, however, there will come a point when the circle
tangent to the base degenerates to a straight line. For alpha larger
still, this circle will contain the other two (and the incircle). At
and above this critical alpha, there is no solution.
Problem 2 (medium): what is this critical alpha?
More generally, if we remove the restriction to isosceles triangles...
Problem 3 (hard): characterize the set of triangles for which an
external kissing configuration exists.
Does anyone have references for this configuration? For example, if
one forms a triangle from the centers of the three circles, then the
incenter of the original triangle would be called the "inner Soddy
center" of the new one -- cf.
http://mathworld.wolfram.com/InnerSoddyCircle.html.
This kind of thing has been compiled by Kimberling
http://faculty.evansville.edu/ck6/encyclopedia/ETC.html,
and often served up less painfully at Mathworld. However, in this
case the existence of external, degenerate, and internal kissing
configurations may render too inelegant any related triangle centers
one might wish to define. (The inner Soddy center notwithstanding,
because it is centers of the original triangle that are sought.)
Jan Kristian Haugland
> Picture a triangle with three circles each of which is externally
> tangent to one of the triangles' sides, and which are mutually
> externally tangent to each other and the triangle's incircle. We will
> call this an "external kissing" configuration for the triangle.
>
> In other words, these circles are the black circles in the "Kissing
> the triangle's incircle is the tiny red circle, and the triangle
> itself must be drawn in without piercing any circles.
>
> Such circles are not always possible. For example, consider an
> isosceles triangle with apex angle alpha. When alpha = pi/3, (i.e.,
> the triangle is equilateral), such circles clearly exist.
>
> Problem 1 (easy): what is radius of these circles (as a multiple of
> the triangle's inradius r)?
>
> As we increase alpha, however, there will come a point when the circle
> tangent to the base degenerates to a straight line. For alpha larger
> still, this circle will contain the other two (and the incircle). At
> and above this critical alpha, there is no solution.
>
> Problem 2 (medium): what is this critical alpha?
>
> More generally, if we remove the restriction to isosceles triangles...
>
> Problem 3 (hard): characterize the set of triangles for which an
> external kissing configuration exists.
>
> Does anyone have references for this configuration? For example, if
> one forms a triangle from the centers of the three circles, then the
> incenter of the original triangle would be called the "inner Soddy
> This kind of thing has been compiled by Kimberlinghttp://faculty.evansville.edu/ck6/encyclopedia/ETC.html,
> and often served up less painfully at Mathworld. However, in this
> case the existence of external, degenerate, and internal kissing
> configurations may render too inelegant any related triangle centers
> one might wish to define. (The inner Soddy center notwithstanding,
> because it is centers of the original triangle that are sought.)
Hi,
I tried to solve it with circle inversion. I haven't gotten to
the bottom of an explicit characterization, but I have found a
parameterization of all such triangles that you might find useful.
Parameters: b satisfies 0 <= b <= 1
a satisfies |a| <= 1 - sqrt(1-b^2)
r satisfies r > 0
The coordinates of the vertices are:
(r/(2-2a), 0)
(-r/(2+2a), 0)
(0, r(b+1)/((b+1)^2 - (a+1)^2))
b = 1 gives the extreme cases when the largest circle becomes a
straight line.
a = 0 gives cases with two equal sides.
Combining them and putting r = 6 (say) gives an answer to Problem 2.
---
J K Haugland
http://www.neutreeko.net
Jan Kristian Haugland
for a = 0).
Basically, from (a, b) we invert the points (-2-a, b), (2-a, b) and a
third point
which is _not_ (a, (a^2 + 2a - b) / (b+1)) as I first thought but
rather a point
(x, y) distinct from (a, b) (except for (a, b) = (0, 0)) satisfying
(i) ((x+1)^2 + ((a^2+2a+b^2)/(2b+2)-y)^2 = (((a+1)^2+(b+1)^2)/(2b+2))
^2
(ii) ((1-x)^2 + ((a^2-2a+b^2)/(2b+2)-y)^2 = (((a-1)^2+(b+1)^2)/(2b+2))
^2
Subtracting (ii) from (i) I got x/(1+y) = a/(1+b). (To be continued?)
Jan Kristian Haugland
(r/(2-2a), 0)
(-r/(2+2a), 0)
(-ra/(a^2+b^2+2b), r(1+b)/(a^2+b^2+2b))
Jake
> Picture a triangle with three circles each of which is externally
> tangent to one of the triangles' sides, and which are mutually
> externally tangent to each other and the triangle's incircle. We will
> call this an "external kissing" configuration for the triangle.
>
> In other words, these circles are the black circles in the "Kissing
> circles" figure at
> http://en.wikipedia.org/wiki/Descartes_theorem, the triangle's incircle
> is the tiny red circle, and the triangle itself must be drawn in without
> piercing any circles.
>
> Such circles are not always possible. For example, consider an
> isosceles triangle with apex angle alpha. When alpha = pi/3, (i.e., the
> triangle is equilateral), such circles clearly exist.
>
> Problem 1 (easy): what is radius of these circles (as a multiple of the
> triangle's inradius r)?
>
> As we increase alpha, however, there will come a point when the circle
> tangent to the base degenerates to a straight line. For alpha larger
> still, this circle will contain the other two (and the incircle). At
> and above this critical alpha, there is no solution.
>
> Problem 2 (medium): what is this critical alpha?
>
> More generally, if we remove the restriction to isosceles triangles...
>
> Problem 3 (hard): characterize the set of triangles for which an
> external kissing configuration exists.
If a<b<c are the three angles of the triangle, then it works if cos(a/2)
* cos(b/2) < cos(c/2). I did it old school:
Let x,y,z be the radii of the three circles (x opposite to a, etc.) and
normalize so that the incircle has radius 1. Consider the triangle formed
by the centers of the three circles near a--that is, the incircle and the
circles of radii y and z. This triangle has sides of lengths 1+y, 1+z,
and y+z, and the angle opposite to the last side is pi-a. The law of
cosines says that:
(y+z)^2 = (1+y)^2 + (1+z)^2 - 2(1+y)(1+z)cos(pi-a)
So:
2(1+y)(1+z)cos(a) = (y+z)^2 - (1+y)^2 - (1+z)^2
2(1+y)(1+z)(cos(a)+1) = (y+z)^2 - (1+y)^2 - (1+z)^2 + 2(1+y)(1+z) = 4yz
cos(a/2)^2 = (cos(a)+1)/2 = y/(1+y) * z/(1+z)
Similarly:
cos(b/2)^2 = x/(1+x) * z/(1+z)
cos(c/2)^2 = x/(1+x) * y/(1+y)
And therefore:
x/(1+x) = cos(b/2) * cos(c/2) / cos(a/2)
y/(1+y) = cos(a/2) * cos(c/2) / cos(b/2)
z/(1+z) = cos(a/2) * cos(b/2) / cos(c/2)
These can be solved for x,y,z only if the right-hand sides are less than
1. If a<b<c, then 0 < cos(c/2) < cos(b/2) < cos(a/2) < 1. So two of the
inequalities are automatic, and the remaining one is cos(a/2) * cos(b/2)
< cos(c/2).
Jan Kristian Haugland
> (-r/(2+2a), 0)
> (-ra/(a^2+b^2+2b), r(1+b)/(a^2+b^2+2b))
I tried this out numerically and arrived
at the following conjecture. A triangle is
admissible if and only if the sum of the
longest side and the shortest height is
less than the sum of the two shortest
sides.
Presumably this is equivalent to Jake's
condition but I just woke up and I will
check it later.
Jim Ferry
> On Fri, 04 Dec 2009 08:47:36 -0800, Jim Ferry wrote:
> > Picture a triangle with three circles each of which is externally
> > tangent to one of the triangles' sides, and which are mutually
> > externally tangent to each other and the triangle's incircle. We will
> > call this an "external kissing" configuration for the triangle.
>
> > In other words, these circles are the black circles in the "Kissing
> > circles" figure at
> >http://en.wikipedia.org/wiki/Descartes_theorem, the triangle's incircle
> > is the tiny red circle, and the triangle itself must be drawn in without
> > piercing any circles.
>
> > Such circles are not always possible. For example, consider an
> > isosceles triangle with apex angle alpha. When alpha = pi/3, (i.e., the
> > triangle is equilateral), such circles clearly exist.
>
> > Problem 1 (easy): what is radius of these circles (as a multiple of the
> > triangle's inradius r)?
Great solution Jake! (see below). I'll go ahead and fill in the
answers
to the special cases of Problems 1 and 2.
Solving for x in Jake's solution, we have
x = 1/(cos(a/2) sec(b/2) sec(c/2) - 1).
For a = b = c = pi/3, this reduces to x = 3 + 2 sqrt(3).
Alternatively, one can apply Descartes' theorem to the x = y = z case
to get
(3/x + 1)^2 = 2 (3/x^2 + 1), which reduces to x^2 - 6x - 3 = 0, for
which
the only positive solution is x = 3 + 2 sqrt(3).
> > As we increase alpha, however, there will come a point when the circle
> > tangent to the base degenerates to a straight line. For alpha larger
> > still, this circle will contain the other two (and the incircle). At
> > and above this critical alpha, there is no solution.
>
> > Problem 2 (medium): what is this critical alpha?
From Jake's solution, we have that the critical alpha satisfies
cos((pi-alpha)/4)^2 = cos(alpha/2),
whence alpha = Arccos(7/25) = 2 Arccos(4/5) ~= 73.74 degrees.
Alternatively, this critical configuration gives one an
opportunity to mention Ford circles (cf. Wikipedia). The
two non-degenerate circles thus have radii equal to 4 times
the incircle, 3-4-5 right triangles abound, and one quickly
finds that alpha = 2 Arccos(4/5).
Very nice! I hadn't solved the general problem when I posed it, and
had
intended to get out my copy of Pedoe (Geometry: A Comprehensive
Course)
to look up how he proved Descartes' Theorem using an "algebra of
circles".
Another way to say state the solution is to let f(x) = log(sec(x/2)):
then
the requirement is that applying f to each angle yields lengths which
form
a triangle.
Anyone up for a generalization?
The generalization of Descartes' Theorem to n dimensions is
straightforward. It doesn't seem to be on Wikipedia, but see
http://mathworld.wolfram.com/DescartesCircleTheorem.html
Therefore we have...
Problem 4: characterize the set of simplices for which an
external kissing configuration exists.
> > Does anyone have references for this configuration? For example, if one
Jan Kristian Haugland
> On Fri, 04 Dec 2009 08:47:36 -0800, Jim Ferry wrote:
> > Picture a triangle with three circles each of which is externally
> > tangent to one of the triangles' sides, and which are mutually
> > externally tangent to each other and the triangle's incircle. We will
> > call this an "external kissing" configuration for the triangle.
> > Problem 3 (hard): characterize the set of triangles for which an
> > external kissing configuration exists.
>
> If a<b<c are the three angles of the triangle, then it works if cos(a/2)
> * cos(b/2) < cos(c/2).
Here is an argument for my earlier claim that an alternative condition
is that the sum of the longest side and the shortest height is less
the
sum of the two shortest sides.
Basically I will give a parametrization of triangles for which
equality
holds, and show that we have cos(a/2) * cos(b/2) = cos(c/2). Then it
remains to check that the inequality goes the right way for other
triangles, but one can just consider an equilateral triangle, for
example.
We start with the parametrization I gave in an earlier post, namely
(r/(2-2a), 0)
(-r/(2+2a), 0)
(-ra/(a^2+b^2+2b), r(1+b)/(a^2+b^2+2b))
We put b = 1 to get the extreme cases, r = 2(1-a^2)(a+b)^2 to get rid
of
fractions, move everything in x-direction so that the third point
meets
the y-axis, and replace the parameter a by t since an angle is
labelled
a. So our triangle has the points
A (-(3-5t+t^2+t^3), 0)
B (3+5t+t^2-t^3, 0)
C (0, 4-4t^2)
where -1 < t < 1.
Also let O denote origo (0, 0).
Side lengths:
AB = AO + OB = 6+2t^2
AC = sqrt(OA^2 + OC^2) = 5-3t-t^2-t^3
BC = sqrt(OB^2 + OC^2) = 5+3t-t^2+t^3
One can check that AB is the longest side, and the height h is 4-4t^2,
so that AB + h = AC + BC.
Let a, b and c denote that angles at A, B and C respectively. It is
clear that c is the largest angle since AB is the longest side.
We have cos(a/2) = sqrt((1+cos a)/2) = sqrt(4/(5+2t+t^2))
and cos(b/2) = sqrt((1+cos b)/2) = sqrt(4/(5-2t+t^2))
Now, cos c = (AC^2 + BC^2 - AB^2) / (2 AB AC)
= (OA^2 + h^2 + OB^2 + h^2 - OA^2 - 2 OA OB - OB^2) / (2 AB AC)
= (h^2 - OA OB) / (AB AC)
= ((4-4t^2)^2 - (3-5t+t^2+t^3)(3+5t+t^2-t^3))/((5-3t-t^2-t^3)(5+3t-
t^2+t^3))
so that
cos(c/2) = sqrt((1+cos c)/2) = sqrt(16/((5+2t+t^2)(5-2t+t^2)))
= cos(a/2) * cos(b/2).
It should be possible to find a parametrization for three dimensions
also,
but I expect it to be messy, so I might put that on ice for a while.
Jim Ferry
wrote:
> On 7 Des, 01:19, Jake <derpif...@yahoo.com> wrote:> On Fri, 04 Dec 2009 08:47:36 -0800, Jim Ferry wrote:
> > > Picture a triangle with three circles each of which is externally
> > > tangent to one of the triangles' sides, and which are mutually
> > > externally tangent to each other and the triangle's incircle. We will
> > > call this an "external kissing" configuration for the triangle.
> (...)
> > > Problem 3 (hard): characterize the set of triangles for which an
> > > external kissing configuration exists.
>
> > If a<b<c are the three angles of the triangle, then it works if cos(a/2)
> > * cos(b/2) < cos(c/2).
>
> (...)
>
> Here is an argument for my earlier claim that an alternative condition
> is that the sum of the longest side and the shortest height is less
> the
> sum of the two shortest sides.
>
>
> so that
>
> cos(c/2) = sqrt((1+cos c)/2) = sqrt(16/((5+2t+t^2)(5-2t+t^2)))
> = cos(a/2) * cos(b/2).
>
> It should be possible to find a parametrization for three dimensions
> also,
> but I expect it to be messy, so I might put that on ice for a while.
That's a pretty neat characterization, Jan. Here are a few comments:
First, let's adopt the more standard notation of a, b, and c being the
side lengths opposite the angles A, B, and C, respectively. We assume
A <= B <= C, or, equivalently, a <= b <= c. Your condition is a + b >
c + h_c, where h_c is the height perpendicular to side c. Letting s =
(a+b+c)/2 be the semiperimeter, and K the area, this is equivalent to a
+b-c > 2K/c, which we may rewrite
(1) c(s-c) > K.
We may also write this as c > r_c, where r_c = K/(s-c) is the radius
of excircle touching side c (this is different from the circles we're
talking about). This is kind of nifty. Anyway, here's my method for
getting c(s-c) > K from Jake's characterization cos(C/2) > cos(A/2)*cos
(B/2):
Let r and R be the radii of the incircle and circumcircle,
respectively. The point where the incircle touches side b, for
example, breaks it into segments of length s-a (touching A) and s-c
(touching C), so tan(C/2) = sin(C/2)/cos(C/2) = r/(s-c). On the other
hand, the law of sines says that 2R = c/sin(C) = c/(2 sin(C/2) cos(C/
2)). Combining these, we find
(2) cos(C/2)^2 = c(s-c)/(4Rr).
Jake's condition involves positive quantities, so it is equivalent to
cos(C/2)^4 > cos(A/2)^2 cos(B/2)^2 cos(C/2)^2, which, using (2) and
its analogs, becomes
c^2(s-c)^2/(4Rr)^2 > abc(s-a)(s-b)(s-c)/(4Rr)^3
Using several well known formulas the L.H.S. becomes
abc(s-a)(s-b)(s-c)/(4Rr)^3 = abcK^2/(4Rrs(4Rr)^2) = abcK^2/(4RK(4Rr)
^2) = K^2/(4Rr)^2,
so c^2(s-c)^2 > K^2, which is just the square of (1).
Jim Ferry
> On Dec 6, 7:19 pm, Jake <derpif...@yahoo.com> wrote:
> > On Fri, 04 Dec 2009 08:47:36 -0800, Jim Ferry wrote:
> > >
> > > Picture a triangle with three circles each of which is externally
> > > tangent to one of the triangles' sides, and which are mutually
> > > externally tangent to each other and the triangle's incircle.
> > > We will call this an "external kissing" configuration for the
> > > triangle.
>
> > > In other words, these circles are the black circles in the "Kissing
> > > circles" figure at http://en.wikipedia.org/wiki/Descartes_theorem,
> > > the triangle's incircle is the tiny red circle, and the triangle
> > > itself must be drawn in without piercing any circles.
> > > Problem 3 (hard): characterize the set of triangles for which an
> > > external kissing configuration exists.
>
> > If a<b<c are the three angles of thetriangle, then it works if cos(a/2)
> > * cos(b/2) < cos(c/2). I did it old school:
>
> > Let x,y,z be the radii of the three circles (x opposite to a, etc.) and
> > normalize so that the incircle has radius 1. Consider thetriangleformed
> > by the centers of the three circles near a--that is, the incircle and the
> > circles of radii y and z. Thistrianglehas sides of lengths 1+y, 1+z,
> > and y+z, and the angle opposite to the last side is pi-a. The law of
> > cosines says that:
>
> > (y+z)^2 = (1+y)^2 + (1+z)^2 - 2(1+y)(1+z)cos(pi-a)
>
> > So:
> > 2(1+y)(1+z)cos(a) = (y+z)^2 - (1+y)^2 - (1+z)^2
> > 2(1+y)(1+z)(cos(a)+1) = (y+z)^2 - (1+y)^2 - (1+z)^2 + 2(1+y)(1+z) = 4yz
> > cos(a/2)^2 = (cos(a)+1)/2 = y/(1+y) * z/(1+z)
>
> > Similarly:
> > cos(b/2)^2 = x/(1+x) * z/(1+z)
> > cos(c/2)^2 = x/(1+x) * y/(1+y)
>
> > And therefore:
> > x/(1+x) = cos(b/2) * cos(c/2) / cos(a/2)
> > y/(1+y) = cos(a/2) * cos(c/2) / cos(b/2)
> > z/(1+z) = cos(a/2) * cos(b/2) / cos(c/2)
>
> > These can be solved for x,y,z only if the right-hand sides are less than
> > 1. If a<b<c, then 0 < cos(c/2) < cos(b/2) < cos(a/2) < 1. So two of the
> > inequalities are automatic, and the remaining one is cos(a/2) * cos(b/2)
> > < cos(c/2).
>
> Anyone up for a generalization?
>
> The generalization of Descartes' Theorem to n dimensions is
> straightforward. It doesn't seem to be on Wikipedia, but see
> http://mathworld.wolfram.com/DescartesCircleTheorem.html
>
> Therefore we have...
>
> Problem 4: characterize the set of simplices for which an
> external kissing configuration exists.
The n-d solution is essentially a generalization of Jake's 2-d
solution. For i = 1 to n+1, let n_i be the outward unit normal vector
for face i (i.e., the face opposite vertex i). Let a_ij denote the
dihedral angle between faces i and j, so that n_i . n_j = -cos(a_ij).
Taking the incenter of the simplex as the origin, we let the inradius
be 1, and let r_i denote the radius of the kissing sphere externally
tangent to face i. The center of this sphere is at c_i = (1 + r_i)
n_i. Therefore, for the given values of n_i, an external kissing
configuration exists iff there are solutions r_i > 0 to the system of
equations
|c_i - c_j| = r_i + r_j, for all 1 <= i < j <= n+1.
Squaring this, we have
(r_i+r_j)^2 = (1+r_i)^2 + (1+r_j)^2 + 2(1+r_i)(1+r_j)cos(a_ij),
which is Jake's equation (where r_i = y, r_j = z, a_ij = a), so it
reduces to the system of equations
cos(a_ij/2)^2 = p_i p_j, for all 1 <= i < j <= n+1, where p_k = r_k/
(1+r_k).
When n > 2, this is an overdetermined set of equations. Its general
solution is
p_i = cos(a_ij/2) cos(a_ik/2) / cos(a_jk/2), for any (i,j,k) which are
all unequal.
An external kissing configuration exists iff for all i from 1 to n+1
we have
(1) p_i is well defined (in that its value is independent of the j and
k used to compute it), and
(2) p_i < 1.
For example, when n = 3, condition (1) reduces to the requirement that
cos(a_12/2) cos(a_34/2) = cos(a_13/2) cos{24/2) = cos(a_14/2) cos
(a_23/2).