Dear all,

Happy New Year.

Eisso wrote:

> while playing around in GSP with various kinds of

> hexagons, I noticed

> that if a hexagon ABCDEF is circumscribed by a conic

> section (i.e. A, B,

> C, D, E, F all lie on the same conic section), then

> the six "short"

> diagonals AC, BD, CE, DF, EA, FB always seem to form

> a hexagon that is

> inscribed by a conic, i.e. the diagonals are all six

> tangent lines to

> one and the same conic.

This problem can be put another way:

Let ABC be a triangle and A1, B1, C1 be points on the

arcs

BC, CA, AB of the ABC circumcircle. Let the distances

of A from the points A1,B1,C1 be a1,a2,a3

of B from the points A1,B1,C1 be b1,b2,b3

of C from the points A1,B1,C1 be c1,c2,c3

Ptolemaeus' theorem gives

aa1 = bb1 + cc1 (1)

bb2 = cc2 + aa2 (2)

cc3 = aa3 + bb3 (3)

If Ab is the intersection (BC, B1C1)= (opp side of A,

opp side of B1)

then from BAb/CAb = area(BA1C1)/area(CA1C1) =

b1b3/c1c3

we conclude that the barycentrics of Ab are

Ab = (0 : c1c3 : b1b3) and similarly

Ac = (0 : c1c2 : b1b2)

Bc = (c1c2 : 0 : a1a2)

Ba = (c2c3 : 0 : a2a3)

Ca = (b2b3 : a2a3 : 0 )

Cb = (b1b3 : a1a3 : 0 )

Finding the equations of the lines BcCb, CaAc, AbBa

and using (1), (2), (3) we prove that these diagonals

of your hexagon are concurrent. Hence by the inverse

Brianchon's theorem we conclude that this hexagon

has an inconic.

Best regards

Nikolaos Dergiades

____________________________________________________________

Do You Yahoo!?

Αποκτήστε τη δωρεάν @... διεύθυνση σας στο

http://www.otenet.gr