ADGEOM 4209 * ADGEOM 4212 * ADGEOM 4213 *

#4209

Dear Geometers,

Let ABC be a triangle, Let two point Ba, Ca lie on BC; two points Ab, Cb lie on AC, two points Ac, Bc lie on AB. Such that six points Ba, Ca, Ab, Cb, Ac, Bc lie on a circle with center (O). Let Oa, Ob, Oc be the centers of three circles (OBaCa), (OCbAb), (OAcBc) respectively.

Let A', B', C' be the points on OOa, OOb, OOc such that OA'=OB'=OC' then AA', BB', CC' are concurrent.

When (O) be the incircle we have de Villiers theorem

When (O) be the circumcircle we have Kosnita theorem

https://www.geogebra.org/m/hfKCbUaP

Best regards

Sincerely
Dao Thanh Oai 

 

 

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#4212

 

Dear Dao Thanh Oai,

 

I dislike the topic "A generalization Kosnita theorem and de Villiers theorem".

You found a nice configuration(7 cycles) on a Möbius plane.

I give the triangle version here and post the correct 7 cycle version later.

second point of intersection

A' - (OCbAb),(OAcBc)

B' - (OAcBc),(OBaCa)

C' - (OCbAb),(OAcBc)

reflection of O in sideline

O'a - BC

O'b - CA

O'c - AB

Then the three circles (O,A',O'a),(O,B',O'b),(O,C',O'c) concur at another point!

Best regards,

Tsihong Lau

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#4213

 

Dear Dao Thanh Oai,

 

Now I post seven cycle version.

Six points A1A2,B1B2,C1C2 are concyclic, P and P' are inverses with respect to the cycle.

Define six cycles as:

cycle - points ; cycle - points

a12 - A1,A2,P ; a'12 - A1,A2,P'

b12 - B1,B2,P ; b'12 - B1,B2,P'

c12 - C1,C2,P ; c'12 - C1,C2,P'

inverse of P - with respect to ; inverse of P' - with respect to

A' - a'12 ; A - a12

B' - b'12 ; B - b12

C' - c'12 ; C - c12

point of intersection - cycles ; point of intersection - cycles

BC - b12,c12 ; BC' - b'12,c'12

CA - c12,a12 ; CA' - c'12,a'12

AB - a12,b12 ; AB' - a'12,b'12

Define six cycles as:

cycle - points ; cycle - points

a34 - A',BC,P ; a'34 - A,BC',P'

b34 - B',CA,P ; b'34 - B,CA',P'

c34 - C',AB,P ; c'34 - C,AB',P'

Then cycles a34,b34,c34 and a'34,b'34,c'34 concur at another point U and U' respectively, which are inverses with respect to the first cycle.

points of intersection - two cycles and the first cycle

A3,A4 - a34,a'34

B3,B4 - b34,b'34

C3,C4 - c34,c'34

If P or P' is the point at infinity, a12,b12,c12 or a'12,b'12,c'12 become a trilateral(triangle).

 

Best regards,

Tsihong Lau