Thursday, April 24, 2014

Centers facts 2


X(37): Let {cA} be the circle with center on (BC) and tangent to (AB) and (AC). {cA} cuts (BC) at Ab, Ac. Build Ba, Bc, Ca, Cb cyclically. Points Ab, Ac, Ba, Bc, Ca, Cb lie on an ellipse with center X(37). [03-13-2014]

X(57): Let A' be the antipode of the A-extouch in the A-excircle and (a') the tangent to the A-excircle through A'. Build (b') and (c') cyclically. The triangle bounded by (a'), (b'), (c') and the triangle ABC are perspective with perspector X(57).

X(100): If A'B'C' is the excentral triangle of ABC, then the Euler lines of triangles A'BC, B'CA and C'AB concur at X(100) [04-23-2014]

X(140): Let A'B'C' and A"B"C" be the medial and the orthic triangle of ABC, respectively. Let A*, B*, C* be the midpoints of [A'A"], [B'B"] and [C'C"], respectively. The Euler lines of AB*C*, BC*A* and CA*B* concur at X(140). Also ABC and A*B*C* are orthologic with centers X(54) and X(5). [01-28-2014]

X(264): Let A'B'C' be the orthic triangle of ABC and H its orthocenter. The perpendicular line to (B'C') through H  cuts (BC) in A". Built B", C" cyclically. Lines (AA"), (BB"),(CC") concur at X(264) [01-22-2014]

X(278): The circle centered at B with radius AC=b cuts (BC) in two points, being Ab the nearest one to C; the circle centered at C with radius AB=c cuts (BC) in two points, being Ac the nearest one to B. The perpendicular to (BC) through Ab cuts (AB) at Ab* and the perpendicular to (BC) through Ac cuts (AC) at Ac*. Let (La) be the line joing Ab* and Ac*. Build (Lb) and (Lc) cyclically. ABC and the triangle bounded by (La), (Lb) and (Lc) are perspective with perspector X(278). (06-11-2013)

X(281): The circle centered at B with radius AC=b cuts (BC) in two points, being Ab the fartest one to C; the circle centered at C with radius AB=c cuts (BC) in two points, being Ac the farthest one to B. The perpendicular to (BC) through Ab cuts (AB) at Ab* and the perpendicular to (BC) through Ac cuts (AC) at Ac*. Let (La) be the line joing Ab* and Ac*. Build (Lb) and (Lc) cyclically. ABC and the triangle bounded by (La), (Lb) and (Lc) are perspective with perspector X(281). (06-11-2013)

X(371): Let O be the circumcenter of ABC. The line through O perpendicular to(AO) cuts the circumcircle ABC at Ab and Ac, where Ab is on the small arc AB and Ac is on the small arc AC. Let A' be the intersection of lines (BAb) and (CAc). Build B', C' cyclically. Lines(AA'), (BB') and (CC') meet at X(371)=Kemnotu point. [02-13-2014]

X(381): Let Ab,Ac be the reflections of A on B and C, respectively. Let A' be the intersection of the diagonals of BAbAcC. Build B',C' cyclically. The center of circle {A'B'C'} is X(381)=midpoint[G,H]. [01-26-2014]

X(479): Let {a'} be the circle through B and C and tangent to the  incircle {i} of ABC. Let A"={a'}/\{i}. Build B" and C" cyclically. Lines (AA"), (BB") and (CC") concur at X(479). [11-29-2013]

X(505): Let A' be the A-excenter of ABC. Let A* be the nearest point to B where line (A'B) cuts the A-excircle and A** the nearest point to C where line (A'C) cuts the A-excircle. Build B*, B**, C*, C** cyclically. The triangle bounded by lines (A*A**), (B*B**) and (C*C**) is perspective to ABC and the perspector is X(505) (3rd isoscelizer point) Same result taking antipodes of A*,A** in A-excircle and so on [05-30-2013]

X(523): Let A'B'C' and A"B"C" be the medial and the orthic triangle of ABC. Through B' and C" are drawn the tangents to the NPC of ABC and they meet at A*. B* and C* are built cyclically. Triangles ABC and A*B*C* have perspector X(523). This point lies on the the line at infinity. Same result is found when the the tangents are drawn through B" and C'. [06-25-2013]

X(631): Let A’B’C’ be the cevian triangle of O in ABC. Let A”,B”,C” be the midpoints of [OA’], [OB’], [OC’], resp. Let A*=(BC”)/\(CB”), B*=(CA”)/\(AC”), C*=(AB”)/\(BA”). A*B*C* and the anticomplementary triangle of ABC are perspective with perspector X(631)=3*OG/5. [03-08-2014]

X(1001): Let A'B'C' and A"B"C" be the medial and excentral triangles of ABC. The lines A"B' and A"C' cut (BC) in A1 and A2. Build B1,B2,C1,C2 cyclically. The points A1, A2, B1, B2, C1, C2 lie on a conic with center X(1001). (06-07-2013)

X(1118): Let A'B'C' be the orthic triangle of ABC. Let A" be the contact point of the A-excircle and its tangent through A'. Build B" and C" cyclically. The perspector of ABC and A"B"C" is X(1118).  

X(1157): Let A'B'C' be the reflection triangle of ABC. Circles {AB'C'}, {BC'A'} and {CA'B'} concur at X(1157) [01-29-2014]

X(1263): Let H and A'B'C' be the orthocenter and orthic triangle of ABC and A"B"C" the reflecion triangle of A'B'C'. Circumcircles of HAA", HBB" and HCC" concur at H and X(1263). [02-11-2014]

X(2051): Let (a) be the line containing the polar of the Spieker center X(10) of ABC in its A-excircle. Build (b) and (c) cyclically. The triangle bounded by the lines (a), (b), (c) is perspective to ABC with perspector X(2051). (06-01-2013)

X(2184): Let (a) be the line containing the polar of the orthocenter X(4) of ABC in its A-excircle. Build (b) and (c) cyclically. The triangle bounded by the lines (a), (b), (c) is perspective to ABC with perspector X(2184). (06-01-2013)

X(2646): Let Ia be the A-excenter of triangle ABC: Denote by Aa, Ab, Ac the tangency points of the A-excircle with the sides (BC), (CA), (AB), respectively. The circumcircles of triangles AIaAa, BIaAb, CIaAc have a common point A", different from Ia. [Problem S167, Issue 4, 2010 – Mathematical Reflections, [www.awesomemath.org/wp-content/uploads/reflections/2010_4/MR4-2010-problems.pdf]. If B" and C" are built cyclically then A"B"C" and the orthic triangle of ABC are perspective with perspector X(2646). [06-28-2013]

X(3062): Let (a'), (b'), (c') be the polars of the incenter in the excircles and A'=(b')/\(c'). Build B'and C' cyclically.  ABC and A'B'C' are perspective with perspector X(3062)

X(3062): Let Ba, Ca be the inverses of B and C in A-excircle. Let (a')=(BaCa) and build (b'), (c') cyclically. ABC and the triangle bounded by (a'), (b'), (c') are perspective with perspector X(3062) [11-21-2013]

X(3062): Let A' be the center of the circle which is tangent to (AB), (AC) and the circumcircle (externally). Let Ab, Ac be the orthogonal projections of A' on (AC), (AB), respectively. Let (a')=(AbAc). Build (b'), (c') cyclically. Then ABC and the triangle bounded by (a'), (b'), (c') have perspector X(3062).
Note: A'=[(1/4)*(-2*a*b*c-b*c^2-c*a^2+a^3-a*c^2+c^3-b*a^2+b^3-a*b^2-c*b^2)/(a*b*c), 1, 1] [01-20-2014] 

X(3068): Let Ba, Ca, C'a, B'a be the vertex of the A-inner-inscribed-square of ABC, with Ba,Ca on (BC), C'a on (AC) and B'a on (AB). Let B"a=(B-C'a)/\(Ba-B'a) and C"a=(C-B'a)/\(Ca-C'a) and (a")=(B"a-C"a). Build (b") and (c") cyclically. ABC and the triangle bounded by (a"), (b"), (c") are perspective with perspector X(3068)  [12-22-2013]

X(3069): Let Ba, Ca, C'a, B'a be the vertex of the A-outer-inscribed-square of ABC, with Ba,Ca on (BC), C'a on (AC) and B'a on (AB). Let B"a=(B-C'a)/\(Ba-B'a) and C"a=(C-B'a)/\(Ca-C'a) and (a")=(B"a-C"a). Build (b") and (c") cyclically. ABC and the triangle bounded by (a"), (b"), (c") are perspective with perspector X(3069)  [12-22-2013]

X(3090): Let A'B'C' be the orthic triangle of ABC, H its orthocenter and A", B", C" the midpoints of [HA'], [HB'], [HC'], respectively. The Euler lines of triangles A"BC, B"CA, C"AB and ABC concur at X(3090) [03-08-2014] 
Antreas Hatzipolakis has pointed me out that this result can be generalized: if A", B", C" 
are such that HA"/HA'=HB"/HB'=HC"/HC'=t then the Euler lines of triangles A"BC, B"CA, C"AB 
concur at Z(t) for all t.

That is true. For any t:
                    Z(t) = (SB*SC*(1-t)+S^2)/a : : (trilinears)
                          =  3*(t-2)*X(2)-2*(t-1)*X(3)
and Z(t) lies on the Euler line of ABC.

Here is a small list of pairs [t,Z(t)] (for |t|<=12 and computed with step=1/60):
[-12, X(3861)], [-8, X(3845)], [-6, X(3843)], [-5, X(3839)], [-4, X(546)], [-3, X(3832)],
[-8/3, X(3858)],[-12/5, X(3856)], [-2, X(381)], [-5/3, X(3854)], [-8/5, X(3857)], [-3/2, X(3855)],
[-4/3, X(3850)], [-1, X(3091)],[-4/5, X(5066)], [-2/3, X(3851)], [-1/2, X(3545)], [-2/5, X(5072)],
[-1/3, X(5068)], [-1/4, X(3544)], [0, X(5)], [1/4, X(5071)], [1/3, X(5056)], [2/5, X(5055)],
[1/2, X(3090)], [2/3, X(1656)], [3/4, X(5067)], [4/5, X(3628)],[1, X(2)], [7/6, X(3533)], [6/5, X(3526)],
[5/4, X(3525)], [4/3, X(140)], [3/2, X(631)], [8/5, X(549)], [5/3, X(3523)],[7/4, X(3524)], [2, X(3)],
[9/4, X(3528)], [7/3, X(3522)], [12/5, X(548)], [5/2, X(376)], [8/3, X(550)], [14/5, X(3534)],
[3, X(20)], [10/3, X(1657)], [7/2, X(3529)], [11/3, X(5059)], [4, X(30)], [14/3, X(5073)], [5, X(3146)],
[6, X(382)],[7, X(3543)], [8, X(3627)], [10, X(3830)], [12, X(3853)]

X(3579): Let A'B'C' be the excentral triangle of ABC. Let Ab, Ac be the midpoints of [A'B] and [A'C], respectively. Build Ba,Bc,Ca,Cb cyclically. The perpendicular bisectors of [AbAc], [BaBc] and [CaCb] meet in X(3579). [08-24-2013]
This result can be generalized: 
if A", B", C" are such that A’Ab/A'B=A’Ac/A'C=B’Bc/B'C=B’Ba/B'A=C’Ca/C'A=C’Cb/C'B=t
then the perpendicular bisectors of [AbAc], [BaBc] and [CaCb] concur at Z(t) for all t.

Z(t) = (SB*b+c*SC-2*S*R)*t-SA*a+SB*b+c*SC-2*S*R : : (trilinears)
      =  (1-t)*X(1)-(2-t)*X(3)
and lies on line (IO) of ABC.

Pairs [t,Z(t)] (for |t|<12 and computed with step=1/60):
[0, X(40)], [1/2, X(3579)], [2/3, X(165)], [1, X(3)], [4/3, X(3576)], [3/2, X(1385)], [2, X(1)], [3, X(1482)]
X(5423): Let {a'} be the A-excircle of ABC and {a"} the circle through B and C and tangent to {a'}. If A'={a'}/\{a"} and B', C' are built cyclically, then (AA'), (BB') and (CC') concur at X(5423) [11-13-2013]

X(5462): Let A'B'C' and A"B"C" be the medial and orthic triangle of ABC. Let (ra) be the radical axis of the NPC's of A'B"C" and A"B'C'. Build (rb) and (rc) cyclically. The lines (ra), (rb), (rc) concur at X(5462) [08-18-2013]

X(5542): Let A'B'C' be the cevian triangle of the Gergonne point Ge in ABC. The line thorugh Ge parallel to (B'C') cuts [A'C'] and [A'B'] at Ba and Ca, respectively. Build Ab, Cb, Ac, Bc cyclically. The point Ba, Ca, Ab, Cb, Ac, Bc are concyclic on a circumference with center X(5542) and radius (1/2)*r*sqrt(r^2+s^2+8*r*R+16*R^2)/(r+4*R) [11-21-2013]

Wednesday, April 9, 2014

Areal centers


Areal center:
Let ΔA’B’C’ and ΔA”B”C” be two triangles inscribed in ΔABC. The areal center of ΔA’B’C’ and ΔA”B”C” is the point S such that the areas of ΔSA’A”, ΔSB’B” and ΔSC’C” are the same.
The next table shows the trilinear center function of the areal centers for every pair of named inscribed triangles.

Triangles
Areal center
EXTOUCH
INCENTRAL
X(1018)
EXTOUCH
INTOUCH
X(100)
EXTOUCH
LEMOINE
(4*a^2+b^2+c^2)*(a-b)*(a-c)/(4*a^2+b^2+c^2+3*a*(b+c))
On line: (662, 4069)
ETC-search 6-9-13 = 2.279224265191568599975
EXTOUCH
MACBEATH
(a-c)*(a-b)*(S^2+SB*SC)/((-a+b+c)*(SB*b+SC*c+2*S*R))
On lines: (162,4551),(1618,2222)
0.04918013762886010149472
EXTOUCH
MEDIAL
X(100)
EXTOUCH
ORTHIC
X(101)
EXTOUCH
STEINER
(b^2-c^2)/(2*a^3-(b^2+c^2)*a-(b+c)*(b-c)^2)
On lines: (1,4041), (65,4705), (75,4086)
-3.910604001388742886439
EXTOUCH
SYMMEDIAL
(b^2+c^2)*(a-b)*(a-c)/(a*(b+c)+b^2+c^2)
On lines: (163,643), (2298,4284)
4.063394297022946848274
EXTOUCH
YFF CONTACT
(b-c)/(2*a^2-a*(b+c)-(b-c)^2)
On lines: (1,650), (2,522), (57,513), (81,3737), (88,1156), (89,4724),
(105,1024), (279,3676), (676,2006), (885,3911), (1002,4893),
(1022,3675), (1026,5376),  (1121,3227), (1638,2826), (3887,4845)
-2.346744861647027810077
INCENTRAL
INTOUCH
X(4551)
INCENTRAL
LEMOINE
(a-b)*(a-c)*(b+c)*(4*a^2+b^2+c^2)/(a*(2*a^2-b^2-3*b*c-c^2))
On lines: (190,3908), (1723,5612)
-0.4731678066924118032593
INCENTRAL
MACBEATH
(a-b)*(a-c)*(b+c)*(S^2+SB*SC)/(a*((b*c+SW)*SA-S^2))
On line: (692,1897)
-0.1848021007747031358220
INCENTRAL
MEDIAL
X(3952)
INCENTRAL
ORTHIC
X(4559)
INCENTRAL
STEINER
(b^2-c^2)/(a*(a^2-b*c))
On lines: (10,514), (80,291), (313,3261), (513,894), (523,594), (741,2372),  (813,2690), (826,1089), (4013,4049), (4589,5380)
0.1903195606998795675211
INCENTRAL
SYMMEDIAL
(a-b)*(a-c)*(b+c)*(b^2+c^2)
On lines:  (11,594), (100,101), (661,3952), (765,5389), (799,4562),
 (1978,3807)
5.999448746196107405213
INCENTRAL
YFF CONTACT
(b^2-c^2)/(a*(a*(b+c)-2*b*c))
On Kiepert hyperbola
On lines: (2,513), (10,661), (76,693), (98,739), (226,4017), (321,523),
 (649,4672), (671,2787), (876,4358), (898,1290), (1647,4444),
(2051,3667), (4010,4080)
0.7387428613738780332255+C2
INTOUCH
LEMOINE
(a-b)*(a-c)*(4*a^2+b^2+c^2)/(4*a^2-3*(b+c)*a+b^2+c^2)
On line: (643,1019)
7.250456452563039901679
INTOUCH
MACBEATH
(a-b)*(a-c)*(SA^2+SB*SC)/(b*SB+c*SC-2*S*R)
On line: (163,1021)
-0.2097107384364576118994
INTOUCH
MEDIAL
X(100)
INTOUCH
ORTHIC
X(109)
INTOUCH
STEINER
(b^2-c^2)/(2*a^3-(b^2+c^2)*a+(b+c)*(b-c)^2)
On lines: (9,661), (210,4705), (312,1577), (2321,4024), (2341,2610)
2.120399798139991690641
INTOUCH
SYMMEDIAL
(a-b)*(a-c)*(b^2+c^2)/(a*(b+c)-(b^2+c^2))
On lines: (101,4040), (662,3737), (666,1026), (831,919), (1581,2664)
-2.263665312530479505429
INTOUCH
YFF CONTACT
(b-c)/(2*a^2-(b+c)*a+(b-c)^2)
On Feuerbach hyperbola
On lines: (1,3309), (7,3667), (8,514), (9,513), (21,1019), (80,2826),
(104,1477), (294,1027), (314,5214), (764,3680), (876,4876),
(900,3254), (1022,1280), (1023,1308), (1025,5377), (1156,2827)
2.308417730794870076935
LEMOINE
MACBEATH
(5*SW-3*SA)*(S^2+SB*SC)*(SA-SB)*(SA-SC)/(a*(SA^2+S^2-4*SB*SC))
-0.1732939223482771784293
LEMOINE
MEDIAL
(a^2-b^2)*(a^2-c^2)*(4*a^2+b^2+c^2)/a
On lines: (2,353), (99,110), (111,5182), (669,1634), (892,4577),
(1383,1992)
1.948852416877363244776
LEMOINE
ORTHIC
a*(a^2-b^2)*(a^2-c^2)*(4*a^2+b^2+c^2)
On lines: (101,3518), (691,827), (1625,3050)
0.4513117194169890968045
LEMOINE
STEINER
(5*SW-3*SA)*(SB-SC)/(a*(2*SW^2-3*SW*SA-3*S^2))
On lines: (523,599), (598,1499)
2.036872078295437158369
LEMOINE
SYMMEDIAL
(b^2+c^2)*(a^2-b^2)*(a^2-c^2)*(b^2+4*a^2+c^2)/(a*(b^2+c^2-2*a^2))
On lines: (826,4576), (892,5466)
-2.254331264696480276624
LEMOINE
YFF CONTACT
(3*SA-5*SW)*(b-c)/(a*((3*SA-5*SW)*a+(3*SB+SW)*b+(3*SC+SW)*c))
2.243404035004790802826
MACBEATH
MEDIAL
(S^2+SB*SC)*(SA-SB)*(SA-SC)/(a*SA)
Anticomplement of X(2972)
On Johnson circumconic

On lines: (2,1972), (4,94), (20,1075), (51,324), (107,110), (112,925),
(162,655), (250,476),  (323,450), (436,1994), (523,1624), (653,3658),
(685,4630), (877,4576), (1301,1302), (1897,4246), (2052,3060),
(3186,4232)
-0.07059630044964227674098
MACBEATH
ORTHIC
a*SA*(S^2+SB*SC)*(SA-SB)*(SA-SC)
On Johnson circumconic
On lines: (3,125), (25,114), (99,107), (110,351), (343,418), (476,930),
(1625,2081), (3265,4576)
-0.2649080012809499161772
MACBEATH
STEINER
(S^2+SB*SC)*(SB-SC)/(a*(SA+SW-6*R^2))
On line (6,2501)
1.734100567063112800206
MACBEATH
SYMMEDIAL
(SA-SB)*(SA-SC)*(SA+SW)*(S^2+SB*SC)/(a*(SA^2-SB*SC))
On line (879,2966)
-0.2984878140101378863865
MACBEATH
YFF CONTACT
(S^2+SB*SC)*(b-c)/(((S^2+SB*SC)*a-(S^2-SA*SC)*b-(S^2-SA*SB)*c)*a)
1.474964369149482119831
MEDIAL
ORTHIC
X(110)
MEDIAL
STEINER
X(5466)
MEDIAL
SYMMEDIAL
X(4576)
MEDIAL
YFF CONTACT
(b-c)/(a*(b+c-2*a))
On circumhyperbola dual of Yff parabola
On lines: (2,514), (7,3676), (75,693), (86,4833), (88,673), (106,675),
(335,4080), (900,903),  (901,927), (918,4945), (1268,3004), (1647,4089), (1797,2989), (4025,4373), (4555,4618), (4615,5468) 
-0.4590961383710460421040
ORTHIC
STEINER
a*(SB-SC)/((SA-SB)*SB+(SA-SC)*SC)
Isogonal conjugate of X(4226)
On Jerabek hyperbola

On lines: (3,512), (4,3566), (6,924), (68,525), (69,523), (71,4079),
(72,4705), (74,3563), (248,2422), (265,690), (526,895),
(826,3519), (868,879), (1176,1510), (1499,4846)
10.35662876830228477958
ORTHIC
SYMMEDIAL
a*(SA+SW)*(SA-SB)*(SA-SC)/SA
Isogonal conjugate of X(4580)
On lines: (6,67), (25,694), (110,112), (162,660), (394,3162),
(427,3051), (468,2211), (647,1624), (648,670), (933,2623),
(1112,3124), (1576,2445)
-0.4198495218937103159533
ORTHIC
YFF CONTACT
a*(b-c)/((SB+SC)*a-SB*b-SC*c)
On lines: (3,649), (63,513), (295,3572), (1790,3733)
7.711263108722297722455
STEINER
SYMMEDIAL
(b^4-c^4)/(a*(a^2*(b^2+c^2)-2*b^2*c^2))
On lines: (2,512), (141,3005), (850,1502), (882,3266)
1.355867776715018247859
STEINER
YFF CONTACT
X(3120)
SYMMEDIAL
YFF CONTACT
(SW+SA)*(b-c)/(a*((2*SA+SB+SC)*a-(SA+SB)*b-(SA+SC)*c))
On lines: (86,3253), (334,876), (1930,2530)
1.672803830975516369642