Some classical theorems of euclidean geometry are actually instances of theorems of inversive geometry.
In many cases, the connection with inversive geometry is indicated by the phrase "line or circle" in the
statement. In some cases, the statement invoves several lines. These can be viewed as ilines through
the point ∞. In such cases, care may be needed to describe conditions on the ilines.
From inversive geometry, we know that three distinct points A,B,C define a unique iline C(A,B,C).
If the points have complex coordinates a,b,c, then P with coordinate p lies on C(A,B,C) if and only if
the crossratio (a,b,c,p) is real. We may also describe the iline as C(a,b,c).
Most of our results can be proved from their euclidean counterparts by mapping a point to ∞
We give the inversive proofs, not because they are simpler than the best proofs in the literature,
but because the euclidean proofs often use one or two diagrams and then claim that all the other
possible configurations have similar proofs. It is seldom clear that all configurations have
been covered, and no analysis is provided to check this.
definition
A family of distinct ilines is concurrent if they all meet in a point.
We begin with a calculation involving (inversive) crossratios.
Suppose that a,b,c,p,q,r are distinct points, and z≠a,b,c, then
H(a,b,c,p,q,r)
= (a,b,r,z)(b,c,p,z)(c,a,q,z)
= (ar)(bp)(cq)/(aq)(br)(cp)
= H(p,q,r,a,b,c).
Also, H(a,b,c,p,q,r) = (a,b,r,q)(b,c,p,q), so is an inversive invariant.
These follow easily from the definition (a,b,r,z) = (ar)(bz)/(az)(br).
lemma 1
Suppose that p,q,r are not on C(a,b,c).
Then C(a,b,r),C(b,c,p),C(c,a,q) are concurrent if and only if H(a,b,c,p,q,r) is real.
When they concur, the common point is z≠a,b,c.
proof
the pivot theorem
Suppose that C(a,b,r),C(b,c,p),C(c,a,q) concur at z ≠ a,b,c.
Then C(a,q,r),C(b,r,p),C(c,p,q) are concurrent.
proof
From the proof of lemma 1, the condition implies that H(a,b,c,p,q,r) is real.
Since H(a,b,c,p,q,r) = H(p,q,r,a,b,c), the latter is also real.
If a,b or c lie on C(p,q,r), then C(a,q,r),C(b,r,p)C(c,q,p) concur at one of p,q,r.
For example, if aεC(p,q,r), then pεC(a,q,r) so the ilines concur at p.
Otherwise, lemma 1 shows that they are concurrent.
miquel's thereom
Suppose that p,q,r are not on C(a,b,c) and
that C(a,b,r),C(b,c,p),C(c,a,q),C(p,q,r) concur at z ≠ a,b,c.
Then C(a,q,r),C(q,c,p),C(r,p,b),C(a,b,c) are concurrent.
proof
As p,q,r are distinct, we may assume that z≠q,r.
Applying the theorem to C(a,b,r),C(b,c,p),C(c,a,q), C(a,q,r),C(b,r,p),C(c,q,p) are concurrent.
Applying the theorem to C(r,a,b),C(a,q,c),C(q,r,p), C(a,b,c),C(r,b,p),C(q,c,p) are concurrent.
Since the concurrences are not at p, they are at the other intersection of C(b,r,p),C(c,q,p).
Note. Some authors refer to our pivot theorem as miquels theorem, and obtain our miquel's
theorem as a corollary.
There is an intruiging connection with menelaus's theorem. The euclidean version says that
if ΔABC is a euclidean triangle and P is on BC, Q on CA and R on AB, then the
points P,Q,R lie on a line if and only if (AR/RB)(BQ/QC)(CP/PA)=1. In complex coordinates,
the condition becomes (ra))(qb)(pc)/(br)(cq)(ap)=1 which rearranges to H(a,b,c,p,q,r)=1. This means
that C(a,q,r),C(b,p,q),C(c,r,p) are concurrent by lemma 1. In fact, we have
the extended form of menelaus's theorem
Suppose that z is not on C(a,b,c), and that p is on C(b,c,z), q on C(c,a,z), r on C(a,b,z).
Then z is on C(p,q,r) if and only if H(a,b,c,p,q,r)=1. Further, if z is on C(p,q,r), then
C(a,q,r),C(b,r,p),C(c,p,q) meet at a point on C(a,b,c).
The first part is proved by applying a mobius transformation taking z to ∞. The second part
follows from miquel's theorem.
In similar fashion, we have
the extended form of ceva's theorem
Suppose that z is not on C(a,b,c), and that p is on C(b,c,z), q on C(c,a,z), r on C(a,b,z).
C(z,a,p),C(z,b,q),C(z,c,r) are concurrent if and only if H(a,b,c,p,q,r)=1. Further, if they concur, then
C(a,q,r),C(b,r,p),C(c,p,q) are concurrent.
The first part can be deduced from the inversive version of menelaus's theorem as usual.
We now show how the simpson line arises in inversive geometry.
lemma 2
Distinct ilines C(a,b,c), C(a,b,d) meet at angles θ=arg(a,b,c,d) and πθ.
C(a,b,c), C(a,b,d) are orthogonal if and only if (c,d,a,b) is purely imaginary.
proof
Given the distinct points a,b,d, there is a mobius transformation sending a,b,d to 0, ∞, 1.
Say c maps to z. Then (c,d,a,b) = (z,1,0,∞) = z. Also, C(a,b,d) maps to the real axis, and C(a,b,c)
to the line through 0 and z. These lines make angles θ=arg(z) and πθ.
The results follow
Suppose that Z is a point on the iline C, and M≠P. Then there is a unique iline C' through
Z and M orthogonal to C. C,C' meet again in a point W which we call the foot of the
Zperpendicular from M to C. When Z=∞, we get the usual foot of the perpendicular
from M to the line C.
wallace's theorem
Suppose that z is not on C(a,b,c) and that m≠z. Let p,q,r be the feet of the zperpendiculars
from m to C(b,c,z),C(c,a,z),C(a,b,z) respectively. Then zεC(p,q,r) if and only if mεC(a,b,c).
The iline C(p,q,r) is called the simpson zline for the ztriangle ABC for the point m.
proof

