TRIANGE CENTRAL CIRCLES

TRIANGE CENTRAL CIRCLES

From the book: Clark Kimberling: TRIANGLE CENTERS AND CENTRAL TRIANGLES (1998)

PERIODIC SEQUENCE

Let fk(n) be the periodic sequence 1,2,3,...k,1,2,3...k,1,2,3...1,2,3,...k,...

fk(2^t) is periodic, t=0,1,2,3....

Example: k =7

1*,2*,3,4*,5,6,7,1*,2,3,4,5,6,7,1,2*,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4*,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1*,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2*,3,4,5,6,7,.....

The numbers with astetisks is the sequence f7(2^t)

Let k be odd

k = 1 : period : 1

k = 3 : period : 1,2

k = 5 : period : 1,2,4,3

k = 7 : period: 1,2,4

k= 9 : period: 1,2,4,8,7,5

k = 11: period: 1,2,4,8,5,10,9,7,3,6

k = 13: period: 1,2,4,8,3,6,12,11,9,5,10,7

k = 15: period: 1,2,4,8

k = 17: period: 1,2,4,8,16,15,13,9

k = 19: period: 1,2,4,8,16,13,7,14,9,18,17,15,11,3,6,12,5,10

k = 21: period: 1,2,4,8,16,11

Let p(k) be the number of the terms of the period (length of the period)

p(k) is the rank of 2 modulo k ie p(k) is the least number x such that 2^x - 1 = 0 (mod k)

For k = 1, the least x such that 2^x - 1 = 0 (mod 1) is 1: 2^1 - 1 = 1 = 1.1

For k = 3, the least x such that 2^x - 1 = 0 (mod 3) is 2: 2^2 - 1 = 3 = 1.3

For k = 5, the least x such that 2^x - 1 = 0 (mod 5) is 4: 2^4 - 1 = 15 = 3.5

For k = 7, the least x such that 2^x - 1 = 0 (mod 7) is 3: 2^3 - 1 = 7 = 1.7

For k = 9, the least x such that 2^x - 1 = 0 (mod 9) is 6: 2^6 - 1 = 63 = 7.9

For k =11, the least x such that 2^x - 1 = 0 (mod 11) is 10: 2^10 - 1 = 1023 = 93.11

For k =13, the least x such that 2^x - 1 = 0 (mod 13) is 12: 2^12 - 1 = 4095 = 315.13

For k =15, the least x such that 2^x - 1 = 0 (mod 15) is 4: 2^4 - 1 = 15 = 1.15

For k =17, the least x such that 2^x - 1 = 0 (mod 17) is 8: 2^8 - 1 = 255 = 15.17

For k =19, the least x such that 2^x - 1 = 0 (mod 19) is 18: 2^18 - 1 = 262144 = 13797.19

For k =21, the least x such that 2^x - 1 = 0 (mod 21) is 6: 2^6 - 1 = 63 = 3.21

......

For k = 1,3,5,7,..... we have the sequence: 1,3,15,7,63,1023,4095,15,255,262144,3......

Let 2^p(k) -1 = M(k).k

M(k): 1,1,3,1,7,93,315,1,15,13797,3,.... = A165781

p(k) = k - 1 ==> k is prime of the sequence : 3, 5, 11, 13, 19, 29, 37, 53, 59, 61, 67, 83, ...... = A001122

____________FIGURATE_________________

Let 123456...k be a regular k-gon.

We start from 1 and move clockwise and put 1 inside the polygon before 2.

From 2 we move clockwise and put 2 inside the polygon before 4.

From 4 we move clockwise and put 4 inside the poygon before 8.

....

We repeat this counting and moving until we reach a number we have reached already. Then we stop since we have a loop.

Examples for k = 3,5,7,9,11,13,15,17,19,21.

CONCENTRIC CIRCLES

Let ABC be a triangle.

Denote: A1, B1, C1 = The NPC centers of NBC, NCA, NAB, resp.

A2, B2 ,C2 = The NPC centers of A1BC, B1CA, C1AB, resp.

A3, B3, C3 = The NPC centers of A2BC, B2CA, C2AB

An, Bn, Cn = The NPC centers of A_n-1BC, B_n-1CA, C_n-1AB.

On = the circumcenter of the triangle AnBnCn.

The points O1, O2, O3, .... On,.... lie on the Euler line (see HERE)

Denote:

O11 = the reflection of O1 in BC

O111 = the reflection of O11 in AN

O112 = the reflection of O111 in BC

Similarly:

O12 = the reflection of O1 in CA

O121 = the reflection of O12 in BN

O122 = the reflection of O121 in CA

and

O13 = the reflection of O1 in AB

O131 = the reflection of O13 in CN

O132 = the reflection of O131 in AB

The circumcenter of O112O122O132 coincides with the cirumcenter of ABC

The same if we take O2:

O21 = the reflection of O2 in BC

O211 = the reflection of O21 in AN

O212 = the reflection of O211 in BC

Similarly:

O22 = the reflection of O2 in CA

O221 = the reflection of O22 in BN

O222 = the reflection of O221 in CA

and

O23 = the reflection of O2 in AB

O231 = the reflection of O23 in CN

O232 = the reflection of O231 in AB

The circumcenter of O212O222O232 coincides with the cirumcenter of ABC

and so on......

The circumcenter of On12On22On32 coincides with the cirumcenter of ABC

So we have a sequence of concentric circles.

Antreas P. Hatzipolakis, 22 June 2016

MAPPINGS OF THE EULER LINE ONTO ITSELF

NPC centers

1. Let ABC be a triangle and N1, N2, N3 = the NPC centers of OBC, OCA, OAB, resp.

The NPC center of N1N2N3 lies on the Euler line of ABC

Let P be a point on the Euler line of ABC and M1, M2, M3 the midpoints of PN1, PN2, PN3, resp.

The NPC center of M1M2M3 lies on the Euler line of ABC. (ie it is the intersection of the Euer lines of ABC and M1M2M3).

2. Let ABC be a triangle and Na, Nb, Nc = the NPC centers of NBC, NCA, NAB, resp.

The circumcenter of NaNbNc lies on the Euler line of ABC

Let P be a point on the Euler line of ABC and Ma, Mb, Mc the midpoints of PNa, PNb, PNc, resp.

The O (circumcenter) of MaMbMc lies on the Euler line of ABC. (ie it is the intersection of the Euler lines of ABC and MaMbMc.)

3. Let ABC be a triangle and Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp.

The circumcenter of NaNbNc lies on the Euler line of ABC

Let P be a point on the Euler line of ABC and Ma, Mb, Mc the midpoints of PNa, PNb, PNc, resp.

The O (circumcenter) of MaMbMc lies on the Euler line of ABC. (ie it is the intersection of the Euler lines of ABC and MaMbMc).

Which other (than N,I) points have that property?

4. Let ABC be a triangle and Na, Nb, Nc = the NPC centers of GBC, GCA, GAB, resp.

The centroid of NaNbNc lies on the Euler line of ABC

Let P be a point on the Euler line of ABC and Ma, Mb, Mc the midpoints of PNa, PNb, PNc, resp.

The G (centroid) of MaMbMc lies on the Euler line of ABC. (ie it is the intersection of the Euler lines of ABC and MaMbMc).

Circumcenters

1. Let ABC be a triangle and Oa, Ob, Oc = the circumcenters of IBC, ICA, IAB, resp.

The circumcenter of OaObOc lies on the Euler line of ABC

Let P be a point on the Euler line of ABC and Ma, Mb, Mc the midpoints of PNa, PNb, PNc, resp.

The O (circumcenter) of MaMbMc lies on the Euler line of ABC. (ie it is the intersection of the Euler lines of ABC and MaMbMc).

2. Let ABC be a triangle and Oa, Ob, Oc = the circumcenters of NBC, NCA, NAB, resp.

The NPC center of OaObOc lies on the Euler line of ABC

Let P be a point on the Euler line of ABC and Ma, Mb, Mc the midpoints of PNa, PNb, PNc, resp.

The NPC center of MaMbMc lies on the Euler line of ABC. (ie it is the intersection of the Euler lines of ABC and MaMbMc).

3. Let ABC be a triangle and Oa, Ob, Oc = the circumcenters of GBC, GCA, GAB, resp.

The centroid of OaObOc lies on the Euler line of ABC

Let P be a point on the Euler line of ABC and Ma, Mb, Mc the midpoints of PNa, PNb, PNc, resp.

The G (centroid) of MaMbMc lies on the Euler line of ABC. (ie it is the intersection of the Euler lines of ABC and MaMbMc).

Antreas P. Hatzipolakis, 3 June 2016

ANOPOLIS PRIMES

Pn = n + Qk, n is a natural number (including 0) ie n belongs to No = {0, 1, 2, 3, ....} with n > 0

where Qk is the smallest number in the set No - {Q1, Q2,.... Qk-1} such that Pn is prime

n = 1

Q1 = the smallest number in the set No - {Q0} = No - {0} such that 1 + Q1 is prime = 1

P1 = 1 + 1 = 2

n = 2

Q2 = the smallest number in the set No - {Q0, Q1} = No - {0, 1}such that 2 + Q2 is prime = 3

P2 = 2 + 3 = 5

n = 3

Q3 = the smallest number in the set No - {Q0, Q1, Q2} = No - {0, 1, 3} such that 3 + Q3 is prime = 2

P3 = 3 + 2 = 5

n = 4

Q4 = the smallest number in the set No - {Q0, Q1, Q2, Q3} = No - {0, 1, 3, 2} such that 4 + Q4 is prime = 7

P4 = 4 + 7 = 11

and so on.

Pn : 2, 5, 5, 11, 11, 11, 11, 17, 17, 23, 23, 23, 23, 29, 29, 37, 37, 37, 37, 37, 37, 47, 47, 47, 47, 53, 53, 59, 59, 59, 59, 59, 67, 67, 67, 67, 73, 73, 79, 79, 83, 83, 83, 83, 89, 89, 97, 97, 97, 97, 97, 97,.....

Anopolis Prime Numbers Sequence:

An = the missing primes from Pn: 3, 7, 13, 19, 31, 41, 43, 61, 71,.............................

Antreas P Hatzipolakis, 27 Dec. 2014

I - CONCURRENT EULER LINES

Let ABC be a triangle and A'B'C' the cevian triangle of P = I (incenter).

Denote:

Oab, Oac = the circumcenters of ABA', ACA', resp.

Oa = the circumcenter of AOabOac. Similarly Ob, Oc

1. The Euler lines of ABC, AOabOac, BObcOba, COcaOcb are concurrent.

2. The Orthocenter of OaObOc is lying on the Euler Line of ABC.

Note:

For A'B'C' = cevian triangle of a point P, the circumcircles of AOabOac, BObcOba, COcaOcb are concurrent at O of ABC.

By Miquel theorem in the quadrilateral (AB,BC,CA,AA'), O is lying on the circumcircle of AOabOac. Similarly O is lying on the circumcircles of BObcOba, COcaOcb

Antreas P. Hatzipolakis, 3 October 2014

*** The Euler lines of ABC, AOabOac, BObcOba, COcaOcb are concurrent in X(186) = Inverse in circumcircle of orthocenter.

*** The Orthocenter of OaObOc is

X=(a^2 (a^2 - b^2 - b c - c^2) (a^5 (b + c) - 2 a^3 (b^3 + c^3) - a^2 b c (b^2 + c^2) + a (b^5 - b^4 c - b c^4 + c^5)+ b c (b^2 - c^2)^2) : ... :...)

lying on the Euler Line of ABC, and with (6-9-13)-search number 1.269044824987441168348286504.

X = (r^2 + 2 r R - R^2 + s^2)X(3) + R^2 X(4)

Angel Montesdeoca. Hyacinthos #22604

1) X(186)

2) a^2 (a^2-b^2-b c-c^2) (a^5 b-2 a^3 b^3+a b^5+a^5 c-a^2 b^3 c-a b^4 c+b^5 c-2 a^3 c^3-a^2 b c^3-2 b^3 c^3-a b c^4+a c^5+b c^5)::

on lines {{2,3},{35,500},{55,5453},{511,5495},{3724,5492}},

Search = 3.7222371386671506172

Agree with barys, however I think the search number is suspect. Reckon it should be 3.7222371386671506172.

Peter Moses

PARALLEL NN-LINES

10. Let ABC be a triangle and P a point.

Denote:

Ab, Ac = the orthogonal projections of A on PB,PC, resp.

Na1 = the NPC center of AAbAc

Na2 = the NPC center of Na1AbAc.

Similarly Nb1, Nb2 and Nc1, Nc2.

The lines Na1Na2, Nb1Nb2, Nc1Nc2 are parallel.

11. If P = I,

the lines Na1Na2, Nb1Nb2, Nc1Nc2 are parallel to Euler Line of ABC

21. Let ABC be a triangle.

Denote:

Na1 = the NPC center of IBC

Na2 = the NPC center of Na1BC.

Similarly Nb1,Nb2, Nc1,Nc2

The lines Na1Na2, Nb1Nb2, Nc1Nc2 are parallel to Euler Line of ABC

31. Let ABC be a triangle and IaIbIc the antipedal triangle of I (excentral triangle)

Denote:

Ab, Ac = the orthogonal projections of A on IaIc, IaIb, resp.

Na1 = the NPC center of AAbAc

Na2 = the NPC center of Na1AbAc

The lines Na1Na2, Nb1Nb2, Nc1Nc2 are parallel to Euler Line of ABC

41. Let ABC be a triangle and IaIbIc the antipedal triangle of I (excentral triangle)

Denote:

Na1 = the NPC center of IaBC

Oa = the circumcenter of IaBC

Nao1 = The NPC center of OaBC.

Similarly Nb1, Nbo1, Nc1, Nco1.

The lines Na1Nao1, Nb1Nbo1, Nc1Nco1 are parallel to OI line of ABC.

Antreas P. Hatzipolakis, 1 October 2014