- In a triangle ABC we inscribe a square BaCaC'aB'a, whose the side BaCa
lies on BC. If r_a is the circumradius of the triangle AB'aC'a, show that:
r_a = ------- = --------- [h_1 = a-altitude; R = circumradius of ABC]
a + h_1 bc + 2aR
If we consider the two other squares inscribed in ABC and the corresponding
Rh_2 Rca Rh_3 Rab
r_b = ------- = ---------, r_c = ------- = ---------
b + h_2 ca + 2bR c + h_3 ab + 2cR
and 1/r_a + 1/r_b + 1/r_c = 3/R + 2(a^2 + b^2 + c^2)/abc.
The three circumcircles are called Lucas Circles.
(I. Panakis: Plane Trigonometry. Athens, 1973. Vol. II, p. 458, #4)
Actually there are two such squares inscribed in the triangle corresponding
to BC (ie whose one side lies on BC), since the vertices Ca, Ba can also lie
on the extensions of AB, AC.
These circles must have some interesting geometrical properties studied by
Lucas. I don't believe that Lucas calculated just their radii and nothing else!
(These three triangles are similar to reference triangle with similitude ratia:
h_1/(a+h_1), h_2/(b+h_2), h_3/(c+h_3), and if I am not mistaken, this was
already known to Hero of Alexandria).
The next exercise in the book above (#5) is on de Longchamps Circle:
A Triangle ABC is given, and we draw the circles (A,a), (B,b), (C,c),
and the circle cutting orthogonally them. Prove that its radius is equal to:
sqrt(2(8R^2 - (a^2+b^2+c^2))
This circle is called de Longchamps Circle.
G. de Longchamps studied the geometrical properties of this circle; not just
its radius! (See the HM thread:
as also JHC did do for the Conway Circle.
So, since the author does not give a reference, I am wondering which are
(Probably, their radical center is a "notable" point, or the triangles
bounded by their common tangents are in perspective with the reference
- Hello Hyacynthians,
Here I continue to attempt to understand via inversive techniques.
In Conway's Global View of inversive triangle geometry it is often
easy to see the existence and properties of circles related to the
triangle. In this global viewpoint we choose any point P on a sphere,
and three equally spaced points on the equator and perform a
stereographic projection, which is inversive and hence angle
preserving, onto the tangent plane on the opposite side of the sphere
from P. By appropriate choice of P we can obtain any triangle, so
that many properties of the triangle can be worked out by observing a
highly symmetric situation on a sphere.
The equator projects to the circumcircle of ABC and the meridian
lines through A, B, and C, which intersect at the north and south
poles, become the Appolonian circles, intersecting in the isodynamic
Aside: The pedal triangle of a point on a Appolonian circle is isosceles.
We see immediately that the Appolonian circles are perpendicular to
the circumcircle. Better yet we can immediately see two more triples
of circles and some of their properties.
On the sphere consider the six points formed by ABC and their
antipodal points A'B'C'. These last points are the second
intersections of the Appolonian meridians with the equator, which
project into the intersection of the Appolonian circles with the
Now consider on the sphere a circle centered at A' going through B
and C. This circle is obviously tangent to the meridians through B
and C. Similar circles can be found at B' and C'. When projected onto
the plane, there are three circles that are each orthogonal to the
circumcircle, go through two vertices of ABC, and are tangent to the
Appolonian circles. Conway calls these the orthogonal circles. More
about them below.
Aside: The pedal triangles of points on orthogonal circles are right.
Now consider on the sphere a circle centered at A and going through
B' and C'. This circle is twice orthogonal to the meridian through A
and tangent to the meridians at B' and C'. Projected into the plane
this means that there is a circle orthogonal to the a-Appolonian
circle and tangent to the other two. This circle is not centered at A
in the plane, because the center of a circle is not preserved under
inversion. There are two other such circles. I do not know the names
of these circles, if indeed they have names.
The orthogonal circles
We know that the exsymmedian lines are the tangents to the
circumcircle at the vertices. Hence the vertices of this triangle
A^K, B^K, C^K are the centers of circles orthogonal to the
circumcircle at two vertices and so are centers of the orthogonal
circles. By symmetry AB^K = CB^K, and since the exsymmedians are
antiparallel the the opposites sides, we conclude that the base angle
of the isosceles triangle on AC is angle B.
Going back to the sphere we find a relevant concurrence in the plane.
Take the meridian great circle through AA' and tilt it so that it
goes through A, A', and P. Note that this goes through the antipodal
point to P. This is also true for the circles PBB' and PCC'. Since
any circle through P projects to a line in the plane, we find that,
in the plane, the the lines AA', BB', and CC' concur. By other
orguments we can determine that the point of concurrence is K, the
symmedian point. Hence as shown in the diagram, B', the intersection
of the b-Appolonian circle with the circumcircle, is on the symmedian
B^K, an exsymmedian point
/ | \
/ | \
/ | \
/ | \
/ | \
/ | \
/ B' \
/ | \
/ <B | <B \
A /-----------|----------\ C
This isoseceles triangle lets us compute the radius CB^K of the
b-orthogonal circle. It equals b/2 . sec B = abc/ (2SB) where SB is
Conway notation for (cc+aa-bb)/2. Calling this radius ra we have a
formula for the three radii of the orthogonal circles
1/ra + 1/rb + 1/rc = 2/abc (SA+SB+SC) = 2S/abc where S = (aa+bb+cc)/2.
The new circles
I am calling them "new" because they are new to me.
The centers of the orthogonal circles were found using the tangent
lines to the circumcircle at A, B, C. The centers of these new
circles are found using the tangents at A', B', C', the three second
intersections of the Appolonian circles with the circumcircle.
These centers are again on the symmedian lines.
This means that ABC, A'B'C', the centers of the orthogonal circles,
and the centers of the new circles are all perspective at K.
For what it is worth some coordinates are
B' = (aa: -bb/2 : cc)
center of orthogonal circle = B^K = (aa : - bb : cc)
center of new circle = (aa : -5bb : cc)
Enough for now.