Let ABC be a triangle and P a point,
Denote:
(Na), (Nb), (Nc) = the NPCs of PBC, PCA, PAB, resp.
D = the Poncelet point of ABCP
The perpendicular to AP through D intersects again (Nb), (Nc) at Ab, Ac, resp.
The perpendicular to BP through D intersects again (Nc), (Na) at Bc, Ba, resp.
The perpendicular to CP through D intersects again (Na), (Nb) at Ca, Cb resp.
Ma, Mb, Mc = the midpoints of AbAc, BcBa, CaCb, resp.
1. Ma, Mb, Mc and D are concyclic.
Center?
2.The perpendicular bisectors of AbAc, BcBa, CaCb, are concurrent at the antipode of D on the circle (MaMbMcD)
Point?
[César Lozada ]:
Again, your configuration leads to very long expressions.
Let’s try it for some particular cases. Denote O*(P), D*(P) the center of the circle and the point of concurrence of the perpendicular bisectors of AbAc, BcBa, CaCb. We have:
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O*(X(1)) = X(13464), circle though ETC’s 11, 3326
D*(X(1)) = REFLECTION OF X(11) IN X(13464)
= (36*sin(A/2)-10*sin(3*A/2))*cos((B-C)/2)+(6*cos(A)-5)*cos(B-C)+16*cos(A)-2*cos(2*A)-15 : : (trilinears)
= a*(2*a^6-5*(b+c)*a^5-(b^2-16*b*c+c^2)*a^4+(b+c)*(10*b^2-23*b*c+10*c^2)*a^3-(4*b^2+19*b*c+4*c^2)*(b-c)^2*a^2-(b^2-c^2)*(b-c)*(5*b^2-13*b*c+5*c^2)*a+(3*b^2-5*b*c+3*c^2)*(b^2-c^2)^2) :: (barys)
= 3*X(1)-X(104), 5*X(1)-X(1768), 9*X(1)-X(12767), 3*X(1)+X(13253), X(80)-3*X(5603), 5*X(104)-3*X(1768), X(104)+3*X(10698), 2*X(104)-3*X(11715), 3*X(104)-X(12767), X(149)-5*X(5734), X(1768)+5*X(10698), 2*X(1768)-5*X(11715), 9*X(1768)-5*X(12767), 3*X(1768)+5*X(13253), 3*X(5603)-2*X(16174)
= lies on the circles (X(1), X(11715)), (X(5), X(15863)) and on these lines: {1, 104}, {4, 7972}, {5, 15863}, {10, 11729}, {11, 11011}, {30, 11274}, {80, 5603}, {100, 7982}, {119, 519}, {145, 12751}, {149, 5734}, {153, 3241}, {214, 517}, {354, 17654}, {515, 1317}, {546, 946}, {551, 6713}, {962, 12119}, {999, 12332}, {1145, 6745}, {1320, 3577}, {1387, 9952}, {1389, 13143}, {1482, 2802}, {1484, 7680}, {2098, 12739}, {2099, 12736}, {2801, 10247}, {2829, 4342}, {3035, 11362}, {3036, 10175}, {3295, 22775}, {3632, 15017}, {3656, 10738}, {4301, 5840}, {4861, 20117}, {5330, 11014}, {5541, 11224}, {5563, 18861}, {5587, 12531}, {5660, 8166}, {5730, 14740}, {5851, 15570}, {5886, 6702}, {5901, 12619}, {6224, 14217}, {6264, 13227}, {6796, 10087}, {6797, 13374}, {7681, 11698}, {8196, 12461}, {8203, 12460}, {9809, 20057}, {9897, 11522}, {10031, 10724}, {10246, 12515}, {10283, 17051}, {10543, 13607}, {10595, 12247}, {11278, 22935}, {11366, 12463}, {11367, 12462}, {11496, 12773}, {11531, 15015}, {12653, 16189}, {12672, 17660}, {22799, 24042}
= midpoint of X(i) and X(j) for these {i,j}: {1, 10698}, {4, 7972}, {100, 7982}, {104, 13253}, {145, 12751}, {962, 12119}, {1317, 1537}, {1320, 6326}, {1482, 6265}, {3244, 21635}, {6224, 14217}, {11278, 22935}, {12672, 17660}
= reflection of X(i) in X(j) for these (i,j): (10, 11729), (11, 13464), (80, 16174), (214, 19907), (5882, 12735), (5884, 5083), (6797, 13374), (10265, 1387), (11362, 3035), (11715, 1), (12619, 5901), (15863, 5), (19914, 6702)
= center of the circle {X(1), X(10698), X(10700)}
= center of the circle {X(100), X(7982), X(8699)}
= X(7972)-of-Euler triangle
= X(10698)-of-anti-Aquila triangle
= X(12295)-of-2nd circumperp triangle
= X(12302)-of-incircle-circles triangle
= X(12893)-of-Hutson intouch triangle
= X(12901)-of-intouch triangle
= X(15863)-of-Johnson triangle
= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (1, 13253, 104), (80, 5603, 16174), (104, 10698, 13253), (2099, 12740, 12736), (5886, 19914, 6702), (6326, 16200, 1320), (10074, 12775, 5450), (10595, 12247, 16173)
= [ -2.9459820625229220, -4.1028699153960670, 7.8407969137306990 ]
----------------------------------
O*(X(6)) =
= 2*a^8-5*(b^2+c^2)*a^6-(3*b^2+c^2)*(b^2+3*c^2)*a^4+(b^2+c^2)*(5*b^4-16*b^2*c^2+5*c^4)*a^2+(b^4-c^4)^2 : : (barys)
D*(X(6) ) =
D*(X(2)) = REFLECTION OF X(115) IN X(8176)
= 4*a^8+16*(b^2+c^2)*a^6-3*(9*b^4+10*b^2*c^2+9*c^4)*a^4+2*(b^2+c^2)*(11*b^4-8*b^2*c^2+11*c^4)*a^2-11*b^8+16*b^6*c^2-18*b^4*c^4+16*b^2*c^6-11*c^8 : : (barys)
= X(5503)+3*X(6054), X(5503)-3*X(9770), 2*X(5569)-3*X(9167), 2*X(7775)+X(14981)
= on lines: {2, 5477}, {4, 543}, {99, 23334}, {114, 524}, {115, 8176}, {325, 2482}, {542, 9756}, {620, 8182}, {1007, 8593}, {1506, 7817}, {2794, 7618}, {3788, 5569}, {5286, 9166}, {5461, 7736}, {6033, 11165}, {6055, 8550}, {6248, 20112}, {7622, 9744}, {9742, 9877}, {9754, 23234}, {18800, 22110}
= midpoint of X(i) and X(j) for these {i,j}: {99, 23334}, {6033, 11165}, {6054, 9770}
= reflection of X(i) in X(j) for these (i,j): (115, 8176), (6055, 9771), (8182, 620)
= X(9880)-of-Artzt triangle
= X(20112)-of-1st anti-Brocard triangle
= [ -6.0987034045326500, -4.8259822220649550, 9.7965152838905690 ]
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O*(X(3) = X(23336) , circle though ETC’s 125, 3258
D*(X(3)) = REFLECTION OF X(125) IN X(23336)
= (SB+SC)*(S^2-3*SB*SC)*((5*R^2-SW)*S^2+(21*R^2-5*SW)*SA^2) : : (barys)
= a^2*(a^10-3*(b^2+c^2)*a^8+(2*b^4+9*b^2*c^2+2*c^4)*a^6+(b^2+c^2)*(2*b^4-9*b^2*c^2+2*c^4)*a^4-(3*b^8+3*c^8+b^2*c^2*(b^4-10*b^2*c^2+c^4))*a^2+(b^4-c^4)*(b^2-c^2)*(b^4+3*b^2*c^2+c^4))*(2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2): : (barys)
= X(26)-3*X(15035), X(265)-3*X(18281), X(7387)-5*X(15040), 5*X(15051)-3*X(18324)
= on lines: {3, 1986}, {26, 15035}, {30, 113}, {49, 17854}, {74, 18445}, {110, 12084}, {125, 12370}, {185, 10226}, {265, 18281}, {1147, 3357}, {2071, 3043}, {2777, 12038}, {3520, 7723}, {5972, 15761}, {6644, 15472}, {6699, 11430}, {7387, 15040}, {10111, 10264}, {10113, 10224}, {10721, 11449}, {12006, 22962}, {12121, 18569}, {12236, 13352}, {12295, 17701}, {12302, 15132}, {12358, 18570}, {12825, 22115}, {12893, 13346}, {13367, 16111}, {13371, 17702}, {15051, 18324}, {15061, 18912}, {15331, 16223}, {15647, 22802}, {16222, 22467}, {18390, 20304}
= midpoint of X(i) and X(j) for these {i,j}: {110, 12084}, {1147, 13293}, {12121, 18569}, {12302, 15132}, {12893, 13346}
= reflection of X(i) in X(j) for these (i,j): (125, 23336), (10113, 10224), (12041, 10226), (15761, 5972), (20773, 1511)
= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (3, 15463, 14708), (1511, 1539, 20771)
= [ 8.2945323984861260, 6.9349372549349660, -4.9886916477334250 ]
----------------------------------
O*(X(6) = MIDPOINT OF X(12039) AND X(19130)
= 2*a^8-5*(b^2+c^2)*a^6-(3*b^2+c^2)*(b^2+3*c^2)*a^4+(b^2+c^2)*(5*b^4-16*b^2*c^2+5*c^4)*a^2+(b^4-c^4)^2 : : (barys)
= 3*X(597)-X(8546), 5*X(3618)-X(8547), 3*X(5476)+X(8542), X(5486)-9*X(14561)
= on lines: {5, 524}, {67, 5133}, {140, 9019}, {373, 468}, {575, 23410}, {597, 1995}, {2854, 10272}, {3618, 8547}, {5486, 14561}, {5943, 15118}, {6593, 13366}, {7495, 9971}, {8262, 13857}, {8550, 12134}, {9976, 12007}, {10301, 19127}, {12039, 19130}
= midpoint of X(12039) and X(19130)
= [ 0.7417055785118562, -0.2876952289829425, 3.4975124503517150 ]
D*(X(6)) = REFLECTION OF X(125) IN O*(X(6))
= a^2*(2*a^12-7*(b^2+c^2)*a^10+(b^4+4*b^2*c^2+c^4)*a^8+(b^2+c^2)*(14*b^4-23*b^2*c^2+14*c^4)*a^6-(8*b^8+8*c^8-3*b^2*c^2*(b^4+6*b^2*c^2+c^4))*a^4-(b^2+c^2)*(7*b^8+7*c^8-b^2*c^2*(23*b^4-38*b^2*c^2+23*c^4))*a^2+(5*b^4-7*b^2*c^2+5*c^4)*(b^4-c^4)^2) : : (barys)
= (SB+SC)*((9*R^2*(18*R^2-13*SW)+SW*(17*SW+3*SA))*S^2-3*(3*R^2*(3*SA-4*SW)-SA^2+SB*SC+2*SW^2)*SA*SW) : : (barys)
= X(5505)+3*X(9970)
= on lines: {6, 12824}, {113, 524}, {576, 2854}, {1495, 6593}, {1511, 9019}, {2781, 4550}, {5505, 9970}, {10752, 11459}
= [ -2.0068605446585200, -3.6551288466253670, 7.0973816271058730 ]
César Lozada
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