### The 5 Circles

"The Problem of Jiang Zemin" (Former Chinese President).
It is said that he used to confront mathematicians as well as school children
with this problem.
Gegeben ist ein nicht-konvexes Fünfeck: fünf Dreiecke ragen
heraus. Die Umkreise benachbarter Dreiecke schneiden sich in zwei
Punkten: einer dieser Schnittpunkte liegt auf zwei der Geraden
- uns interessiert jeweils
der zusätzliche Schnittpunkt: Insgesamt erhalten wir fünf
solche Schnittpunkte und es gilt: **Sie liegen auf einem Kreis.**

## The General Result: Lines in the Plane

**Theorem.**
Let L be a finite set of lines in the plane of cardinality |L| at least 2
and assume that they are in general position.
- If |L| is even, these lines determine a point P(L),
- If |L| is odd, they determine a circle C(L),

with the following property:
- For |L| = 2, P(L) is the intersection of the two lines.

Assume now that |L| is at least 3 and that L' is obtained from L by deleting one line. Then:
- If |L| is even, then C(L') passes through P(L).
- If |L| is odd, then P(L') lies on C(L).

Of course, one can use the properties in order to construct inductively
P(L) and C(L), respectively; in particular P(L) and C(L) are uniquely determined.

**The case n=5:** The pentagon picture above exhibits some of the circles and points:
given
is a set L of five lines,
- the red circle is C(L),
- the black circles are some of the circles C(L''), where L'' is a
triple of lines.
- the intersection of the red circle with the black ones are the points
P(L'), with L' being a quadruple of lines.
- (One could add another 5 black circles by looking at the remaining triples
of lines, any one of these would pass through two of the points P(L').)

References: W.K.Clifford: *Synthetic proof of Miquel's theorem.*
in Mathematical Papers, Macmillan 1882.

I. Yaglom: *Complex Numbers in Geometry.* Academic Press 1968.

In particular, see:
Probleme des 5 cercles
(with a link to a file (ps and pdf) with full proofs, by Thomas Chomette).

See also Clifford's Circle Theorem