>>Dear Hyacinthians,

Precisely, the three pairwise radical axis of the 3 circles , i.e. the

>>

>>The following theorem is an example of a

>>non-standard triangle geometry result because

>>the proof (at least, the proof I know) uses a

>>reductio ad absurdum with the help of the

>>fact that the three pairwise radical axes of

>>three circles always concur:

>>

>> If ABC is a non-degenerate triangle, X and X'

>> are two points on the line BC, Y and Y' are

>> two points on the line CA, and Z and Z' are

>> two points on the line AB, such that the

>> points Y, Y', Z and Z' lie on one circle, the

>> points Z, Z', X and X' lie on one circle, and

>> the points X, X', Y and Y' lie on one circle,

>> then all six points X, X', Y, Y', Z and Z'

>> lie on one circle.

>>

>>I remember having seen this result in a paper

>>by Third in the Edinburgh Math. Proceedings.

>>Can anybody give an *exact* reference with

>>year, page nos, etc.?

>>

>> Thanks a lot!

>> Sincerely,

>> Darij Grinberg

>>

>>

>

>That result looks false to me. Start with distinct overlapping

>circles c1, c2 and c3, and define {X,X'} = c2 /\ c3, {Y, Y'} =

>c3 /\ c1, {Z, Z'} = c1 /\ c2. Then take BC = line XX' etc.

>

>

sidelines of ABC, are concurrant, this fact is impossible for ABC

non-degenerate.

Y, Y', Z, Z' lie on c1

Z, Z', X, X' lie on c2

X, X', Y, Y' lie on c3

The common power of A wrt c1 and c3 is AY.AY'

The common power of A wrt c1 and c2 is AZ.AZ'

Hence A as the same power wrt c2 and c3. But c2/\c3 = {X,X'} and A don't

lie on the line XX' = BC (since ABC is non-degenerate).

So the locus of the points with the same power wrt c2 and c3 is not a

line and c2 = c3 and X, X', Y, Y', Z, Z' lie on the circle c1=c2=c3.

Friendly

Gilles Boutte- Dear Gilles,

In Hyacinthos message #9794, you wrote:

>> Precisely, the three pairwise radical axis of the 3

Yes, this is the proof of Third's theorem.

>> circles , i.e. the sidelines of ABC, are concurrant,

>> this fact is impossible for ABC non-degenerate.

By the way, if I remember correctly, Third formulated his

theorem with the help of antiparallels.

Sincerely,

Darij Grinberg >Dear Hyacinthians,

Was Third's reference earlier than 1949? I remember striking

>

>The following theorem is an example of a

>non-standard triangle geometry result because

>the proof (at least, the proof I know) uses a

>reductio ad absurdum with the help of the

>fact that the three pairwise radical axes of

>three circles always concur:

>

> If ABC is a non-degenerate triangle, X and X'

> are two points on the line BC, Y and Y' are

> two points on the line CA, and Z and Z' are

> two points on the line AB, such that the

> points Y, Y', Z and Z' lie on one circle, the

> points Z, Z', X and X' lie on one circle, and

> the points X, X', Y and Y' lie on one circle,

> then all six points X, X', Y, Y', Z and Z'

> lie on one circle.

>

>I remember having seen this result in a paper

>by Third in the Edinburgh Math. Proceedings.

>Can anybody give an *exact* reference with

>year, page nos, etc.? ....

this problem as part of a longer proof of an old text-book exercise:

E. A. Maxwell, "Geometry for Advanced Pupils," (1949), p.170, ex.14.

The first stage of the proof I know is to show that certain specially

defined points like your Y, Y', Z, Z' are concyclic; which prepares

the way for the radical axes argument which you mention. Maxwell

ascribes his problem to the Oxford and Cambridge Schools Examination

Board, presumably some time before 1949.

Ken Pledger.- Dear Ken Pledger,

Thanks for the information in Hyacinthos message

#9836. Lately, I have succeeded to localise the

Third reference:

J. A. Third, "Systems of circles analogous to

Tucker circles", Proc. Edinburgh Math. Soc.,

17 (1898) p. 70-99.

In fact, the fact I called "Third theorem" is

mentioned and proved on the very first page of

the above article. Third states it using the

notion of antiparallels (instead of saying that

the points Y, Y', Z and Z' lie on one circle,

he says that the lines YZ' and Y'Z are

antiparallel to each other with respect to the

angle CAB).

Thanks to all for the replies to my inquiry.

Sincerely,

Darij Grinberg