- Dear Darij,
>

The theorem of 3 circles states that their homothety centers

> This is very interesting: you state that the

> external homothetic center of the incircles of

> triangles PAB and PBC coincides with the external

> homothetic center of the incircles of triangles

> PCD and PDA. My sketches confirm this, but do you

> have a simple proof?

( external ones, and the circles have to be outside of

each other) are collinear. I do not know who's name this

theorem carries ( no doubt some French guy), but with all this

excess homothety the result of Menelaus theorem probably was used

to prove it.

The diagonals of the quadrilateral keep the entire groop

of 4 circles nicely together, being common tangents.

> >> When you complete the ABCD quadrilateral

You are right. I do not know what I had in mind.

> >> also, the 2 new vertices will be on this line

> >> too.

>

> Not on my figure!

>

You probably realized that I'm not a professional mathematician,

> >> In fact, the homothety centers of the 4

> >> circles together with the point P make an

> >> auto-polar triangle. The orthocenter of this

> >> triangle is outside of it, and is the center

> >> of the circle in question. It is on the line

> >> connecting the P and the incircle O.

>

> Unfortunately, I have lost the thread in these

> investigations, but my dynamic sketches seem to

> contradict this. In fact, in "very obtuse"

> circumscribable quadrilaterals ABCD, the center

> of the circle doesn't lie on PO.

>

> It seems that I am misunderstanding something.

> Can you please elaborate more on the

> explanation of the results?

>

just an amature, who likes math. I'm sorry, I do not always have

rigorous proofs to the statements I make.

I should have taken the sketch to the 'extreme', you are right,

some lines only looked parallel.

But still, the perpendiculars from the homothety centers on the

continuations of the diagonals onto the other diagonal do cross

in the center of the resulting circle , being the orthocenter

of the auto-polar triangle.

Another point: if you take a Gergonne line and rotate it a little

around it's originating vertex, the distance its foot travels along

the opposite side of the triangle is congruent to the distance

that separates now the tangent points of the pair of incircles

on the new position of the 'Gergonne' line. The proof should

be similar to the one that helped M. Pitot to prove his theorem

a+c=b+d for the quadrilateral with a circle inside.

Are you sure there is no very simple solution using angle

property of inscribed quadrilaterals A+C=B+D=2d? There must exist

some pair of pivotal angles that affects all others when the

nice square being twisted into irregular quadrilateral squeezing

without crushing the very round egg inside?

Sorry for the nonsense, old, present, and future!

Rafi. - Dear friends,

I didn't follow your messages so excuse me for repeating the following:

it is known that AI*AJ = AB*AC where I, J are the incenter and A_excenter

of ABC

If the quadrilateral ABCD is circumscribed to a circle P is the intersection

of the diagonals

and I1, I2, I3, I4 are the incenters and J1, J2, J3, J4 are the

P_excenters of the triangles

PAB, PBC, PCD, PDA then since

PI1*PJ1 = PA*PB

PI3*PJ3 = PC*PD

PI2*PJ2 = PB*PC

PI4*PJ4 = PD*PA

we get

PI1*PI3*PJ1*PJ3 = PI2*PI4*PJ2*PJ4

and since

PI1*PI3 = PI2*PI4

we get

PJ1*PJ3 = PJ2*PJ4

or that the excenters are also cyclic points.

Best regards

Nikolaos Dergiades