- In the Greek mathematical periodical Diastasis [=dimension] #3-4, 1998,

pp. 24-37, I read the article by Nikos Kyriazis:

A New Geometry Theorem and its more Significant Applications.

The author uses the theorem as a lemma to prove well-known theorems:

The three angle-bisectors/medians/altitudes of a triangle concur, etc.

Actually the theorem is not new. The earliest reference I found is the

following:

Soit A_1A_2A_3A_4A_5A_6 un hexagone convexe inscrit dans un cercle.

Si le produit A_1A_2 X A_3A_4 X A_5A_6 des cotes de rang pair egale

le produit des cotes de rang impair A_2A_3 X A_4A_5 X A_6A_1, les

diagonales A_1A_4, A_2A_5, A_3A_6 sont concourantes, et reciprocuement.

(L'education mathematique 49(1946-1947) 150, #8941)

Does anyone know an earlier one?

Note that the theorem was once proposed as a problem by HSMC:

Given six consecutive points A, B, C, D, E, and F on a circle, prove

that if (AB)(CD)(EF) = (BC)(DE)(FA), then AD, BE, and CF are concurrent.

(The Mathematics Student Journal, v. 27, #5, 1980, p. 3, # 527,

by H. S. M. Coxeter)

Antreas - Is it possible that there is any connexion with Martin Gardner, The

Asymmetric Propeller, Coll. Math. J., 30(1999) 18-22, and the Bankoff,

Erd"os, Klamkin Math Mag article (1977?) referred to therein? R.

On Sat, 1 Jul 2000 xpolakis@... wrote:

> In the Greek mathematical periodical Diastasis [=dimension] #3-4, 1998,

> pp. 24-37, I read the article by Nikos Kyriazis:

> A New Geometry Theorem and its more Significant Applications.

>

> The author uses the theorem as a lemma to prove well-known theorems:

> The three angle-bisectors/medians/altitudes of a triangle concur, etc.

>

> Actually the theorem is not new. The earliest reference I found is the

> following:

>

> Soit A_1A_2A_3A_4A_5A_6 un hexagone convexe inscrit dans un cercle.

> Si le produit A_1A_2 X A_3A_4 X A_5A_6 des cotes de rang pair egale

> le produit des cotes de rang impair A_2A_3 X A_4A_5 X A_6A_1, les

> diagonales A_1A_4, A_2A_5, A_3A_6 sont concourantes, et reciprocuement.

> (L'education mathematique 49(1946-1947) 150, #8941)

>

> Does anyone know an earlier one?

>

> Note that the theorem was once proposed as a problem by HSMC:

> Given six consecutive points A, B, C, D, E, and F on a circle, prove

> that if (AB)(CD)(EF) = (BC)(DE)(FA), then AD, BE, and CF are concurrent.

> (The Mathematics Student Journal, v. 27, #5, 1980, p. 3, # 527,

> by H. S. M. Coxeter) - On Sat, 1 Jul 2000 xpolakis@... wrote:

> In the Greek mathematical periodical Diastasis [=dimension] #3-4, 1998,

I don't know of an earlier reference but I am glad you raise the

> pp. 24-37, I read the article by Nikos Kyriazis:

> A New Geometry Theorem and its more Significant Applications.

>

> The author uses the theorem as a lemma to prove well-known theorems:

> The three angle-bisectors/medians/altitudes of a triangle concur, etc.

>

> Actually the theorem is not new. The earliest reference I found is the

> following:

>

> Soit A_1A_2A_3A_4A_5A_6 un hexagone convexe inscrit dans un cercle.

> Si le produit A_1A_2 X A_3A_4 X A_5A_6 des cotes de rang pair egale

> le produit des cotes de rang impair A_2A_3 X A_4A_5 X A_6A_1, les

> diagonales A_1A_4, A_2A_5, A_3A_6 sont concourantes, et reciprocuement.

> (L'education mathematique 49(1946-1947) 150, #8941)

>

> Does anyone know an earlier one?

>

> Note that the theorem was once proposed as a problem by HSMC:

> Given six consecutive points A, B, C, D, E, and F on a circle, prove

> that if (AB)(CD)(EF) = (BC)(DE)(FA), then AD, BE, and CF are concurrent.

> (The Mathematics Student Journal, v. 27, #5, 1980, p. 3, # 527,

> by H. S. M. Coxeter)

>

>

point. I do, however, have two references different from yours.

I will write about this when I check my notes at home. For

now I just remember that I once saw the theorem as an excersize

for traing students for a mathematical competition.

I have seen the article of Kyriazis you quote, and I knew the

result years now.

Unfortunately Kyriazis presents it as if he discovered THE theorem

in geometry. His proof is too elaborate but here is a trivial one:

By the trigonometric form of Ceva's theorem applied to triangle

ACE we have

sin(EAD)

-------- .--------.-------- = 1

sin(DAC)

Use know ED=2Rsin(EAD) etc and you are done.The converse similar.

Michael