Monday, February 16, 2015

Tangent circles to circumcircle, NPC and incircle (with extraversions).


There are two circles that are tangent to the circumcircle, NPC and incircle of a triangle ABC. The first one, ΓI, is internally tangent to NPC; the other, ΓE, is externally tangent to NPC.

NOTE: All coordinates are barycentrics.


1.    Internal tangent circle ΓI:

Radius= r*|(r^2-s^2+5*r*R+4*R^2)/(4*r*R+4*R^2+3*r^2-s^2)|

Center Zi:
Zi = 2*a^6-2*(b+c)*a^5-(b^2-4*b*c+c^2)*a^4+2*(b^2-c^2)*(b-c)*a^3+4*(b^2-c^2)*(b-c)*a*b*c-2*(b^2+3*b*c+c^2)*(b-c)^2*a^2+(b^2-c^2)^2*(b-c)^2 : :

= Complement of X(2968)

= Midpoint of: (1897,2968)                                                                        

= On lines:  (1,5), (2,1897), (3,108), (30,1785), (105,5020), (165,1360), (521,3042), (676,2804), (867,1862), (1068,3149), (1532,1870), (1936,5762), (2635,6357)

= (4*R^2-SW)*X(1)+3*r^2*X(2)-r^2*X(3)

= ( 1.611730809126064, 1.55156088792320, 1.822631186056290 )

Passes through X(11) y X(108), having antipodes:

    Reflection of X(11) in Zi :

    X = 2*a^6-2*(b+c)*a^5-(b^2-4*b*c+c^2)*a^4+2*(b^2-c^2)*(b-c)*a^3+4*(b^2-c^2)*(b-c)*a*b*c-2*(b^2+3*b*c+c^2)*(b-c)^2*a^2+(b^2-c^2)^2*(b-c)^2 : :

       = Midpoint of: (100,1897)                                                                         

       = Reflection of: (2968/3035)                                                                        

      = On lines: (1,5), (100,108), (109,900), (523,2222), (2149,3700), (2968,3035)

      = ( 0.580326497006925, -0.26962876949254, 3.559487554476018 )

     Reflection of X(108) in Zi:
X=(b-c)^2*(-b-c+a)*(-b^2-c^2+a^2)*(a^10+(3*b*c-c^2-b^2)*a^8-7*b*c*(b+c)*a^7+(-2*c^4+22*b^2*c^2-2*b^4+b*c^3+b^3*c)*a^6+b*c*(b+c)*(9*c^2-22*b*c+9*b^2)*a^5+(2*b^4-5*b^3*c-18*b^2*c^2-5*b*c^3+2*c^4)*(b-c)^2*a^4+3*(b^2-c^2)*(b-c)*b*c*a^3*(c^2+6*b*c+b^2)+(b^2-c^2)^2*a^2*(c^2+6*b*c+b^2)*(-3*b*c+c^2+b^2)-5*(b^2-c^2)^3*(b-c)*a*b*c-(b^2-c^2)^4*(b-c)^2) : :

= Reflection of: (2968/123)                                                                         

= On lines:  (3,108), (11,123), (1364,2850)

= ( 3.305222351136989, 2.97411313557093, 0.056175841372050 )


2.    External tangent circle ΓE

Radius = r*|(r^2-s^2+5*r*R+4*R^2)/(3*r^2-s^2+12*r*R)|

Center ZE:
ZE = 2*a^6-2*(b+c)*a^5-(b^2-4*b*c+c^2)*a^4+2*(b^2-c^2)*(b-c)*a^3-2*(b^2-b*c+c^2)*(b-c)^2*a^2-4*(b^2-c^2)*(b-c)*a*b*c+(b^2-c^2)^2*(b-c)^2 : :

= On lines:  (1,5), (3,105), (108,1598), (238,5762), (516,3246), (517,3008), (676,2826),  (948,999),                             (1360,3361), (4310,5779)

= SW*X(1)-3*(4*R*r+r^2)*X(2)+(4*R*r+r^2)*X(3)

= ( 4.948911141596107, 7.44414659365719, -3.797088302130342 )

Passes through X(11) y X(105), having antipodes:

    Reflection of X(11) in ZE :
       X =2*a^9-4*(b+c)*a^8+12*a^7*b*c+(b+c)*(5*c^2-14*b*c+5*b^2)*a^6+(-5*c^4+12*b^2*c^2-5*b^4)*a^5+(b^2-c^2)*(b-c)*a^4*(b+2*c)*(2*b+c)+2*(b^4-3*b^3*c+b^2*c^2-3*b*c^3+c^4)*(b-c)^2*a^3-(b^2-c^2)*(b-c)^3*a^2*(3*c^2-2*b*c+3*b^2)+(b^4-c^4)*(b^2-c^2)*a*(b-c)^2-(b^2-c^2)^3*(b-c)*b*c : :

        = On lines: (1,5), (104,105), (106,676), (1320,2398)

        = ( 7.254687161947005, 11.51554264197544, -7.679951421897236 )

    Reflection of X(105) in ZE:
X =(b-c)^2*(a^7-3*(b+c)*a^6+(5*b^2+9*b*c+5*c^2)*a^5-(b+c)*(7*c^2+2*b*c+7*b^2)*a^4+(7*b^4+4*b^3*c+4*b^2*c^2+4*b*c^3+7*c^4)*a^3-(b+c)*(5*c^4-6*b*c^3+6*b^2*c^2-6*b^3*c+5*b^4)*a^2+(3*c^2+4*b*c+3*b^2)*(b^2-b*c+c^2)*(b-c)^2*a-(b^4-c^4)*(b-c)^3) : :

= On lines:  (3,105), (11,1111), (1086,2820)

= ( 7.203893096245062, 5.53877424438016, -3.518745270161146 )



3.    Similitude centers of circles ΓI, ΓE

Insimil-center = X(11)

Exsimil-center
= (2*a^4-2*(b+c)*a^3+(b^2+c^2)*a^2-(b^2-c^2)^2)*(a-b+c)*(a+b-c) :  :

= On lines: (1,5), (25,105), (109,1086), (226,1386), (1331,5856), (1458,6357), (1465,3011), (2361,5762)

= (2*R^2-SW)*X(1)+2*(2*R*r+r^2)*X(5)

= ( -1.089194017590830, -3.21756437369972, 6.370913979510722 )  



4.    Extraversions

Taking A-excircle instead of incircle, we find also two tangent circles to A-excircle, circumcircle and NPC. One of them, ΓAI is internally tangent to NPC, and the other,  ΓAE, is externally tangent to NPC. Let OAI and OAE be their centers and TAI and TAE their points of contact with the circumcircle. Build and name other circles cyclically.  Then:

·         ΔOAIOBIOCI is perspective to EXCENTRAL and FEUERBACH triangles at X(5)

·         ΔOAE OBE OCE is perspective to EXCENTRAL and FEUERBACH triangles at X(5) and it is also perspective to MEDIAL triangle at:

Z = (2*a^6+2*(b+c)*a^5-(b^2-4*b*c+c^2)*a^4-2*(b^2-c^2)*(b-c)*a^3-4*(b^2-c^2)*
(b-c)*a*b*c-2*(b^2+3*b*c+c^2)*(b-c)^2*a^2+(b^2-c^2)^2*(b-c)^2)*(a-b-c)^2
= Complement of X(6356)
 = On lines: (2,1119), (3,281), (5,9), (268,3560), (284,1146), (610,5787), (2322,2968), (3739,5745)
 = ( 4.713912920783893, 3.19721412780796, -0.748443570013695 )

·         Lines (TAI TAE), (TBI TBE), (TCI TCE) concur at X(25)

·         ΔTAITBITCI is perspective to CIRCUMPERP1 and FEUERBACH triangles at X(4220) and X(2), respectively.

·         ΔTAETBETCE is perspective to CIRCUMPERP2 and MIDARC triangles at X(28), to FEUERBACH triangle at X(4), and to TANGENTIAL triangle at:

Z= (-4*cos((B-C)/2)*sin(A/2)*cos(A)-cos(B-C)+2*cos(A)^2+cos(A))*sin(A)^2*sec(A)::
 = On lines: (1,19), (3,281), (4,198), (25,1604), (34,2270), (56,1249), (57,2331), (92,1817), (104,1436), (108,393), (579,1783), (910,1841), (1033,1617), (1400,2202), (1753,2324), (1826,4219), (1870,2262), (1880,1951), (2322,2975), (4224,5089)
 = ( -68.607417041979430, -92.43074522778095, 99.296142120515550 )

5.    Index of centers.

6-9-13 first coordinate
Point
-68.60741704197940
Perspector ΔTAETBETCE - TANGENTIAL triangle of ABC
-1.08919401759083
Exsimicenter of ΓI,  ΓE
0.58032649700693
Reflection of X(11) in ZI
1.61173080912606
Center of ΓI = ZI
3.30522235113698
Reflection of X(108) in ZI
4.71391292078389
Perspector ΔOAEOBEOCE - MEDIAL triangle of ABC
4.94891114159610
Center of ΓE = ZE
7.20389309624506
Reflection of X(105) in ZE
7.25468716194700
Reflection of X(11) in ZE


César Lozada
Feb 16, 2015



No comments:

Post a Comment