X3333

POHOATA POINT

Let R be the circumradius, r the inradius, and \(r_{A}\) the radius of the A-excircle. The trilinears given above are equivalent to 2R - \(r_{A}\) : 2R - \(r_{B}\) : 2R - \(r_{C}\). The 1st trilinear representation
< A B C >: Reference Triangle △ABC [ A B ]: Sideline AB (Base c = 1) [ B C ]: Sideline BC (Side a) [ C A ]: Sideline CA (Side b) { 1, 3339 }: OI Line (Incenter–Circumcenter Line) { 1, 3 }: OI Line (Incenter–Circumcenter Line) { 7, 84 }: Central Line L(7, 84) { 200, 474 }: Central Line L(200, 474) { 10, 1056 }: Central Line L(10, 1056) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_1 = INCENTER (0.5714, 0.2333) X_3339 = INTERSECTION OF LINES X(1)X(3) AND X(7)X(10) (0.6083, 0.3010) X_3 = CIRCUMCENTER (0.5000, 0.1021) X_7 = GERGONNE POINT (0.5903, 0.2769) X_84 = ISOGONAL CONJUGATE OF X(40) (0.6071, 0.6816) X_200 = -CEVA CONJUGATE OF X(9) (0.4392, 0.1037) X_474 = EULER LINE INTERCEPT OF LINE X(10)X(56) (0.5441, 0.2176) X_10 = SPIEKER CENTER (0.5204, 0.1833) X_1056 = POINT ALCYONE (0.5756, 0.2499) X_3333 = POHOATA POINT (0.5899, 0.2671) A B C X1 X3339 X3 X7 X84 X200 X474 X10 X1056 X3333 (a, b, c) = (5, 6, 7) | AB = 1

X3333 (POHOATA POINT) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

Trilinears: \((a - b + c)(a + b - c)(a + b + c) + 4abc : :\)
Search Value (6, 9, 13): \(1.074954505259\)
▸ Incident Central Lines27 lines

{1, 3}, {4, 1435}, {7, 84}, {9, 1125}, {10, 1056}, {48, 1449}, {58, 2191}, {81, 2360}, {109, 1497}, {200, 474}, {226, 3086}, {386, 1066}, {388, 1210}, {495, 1698}, {496, 1699}, {515, 938}, {516, 1058}, {518, 936}, {519, 1706}, {580, 1471}, {581, 1458}, {614, 1453}, {1106, 2263}, {1203, 3157}, {1387, 1768}, {1731, 2257}, {2136, 3244}

▸ Functional Relations3 relations

Notes

Let I be the incenter, X1, and let KA be the symmedian point of triangle IBC; define KB, KC cyclically. Let X be the midpoint of segment AI,and define Y, Z cyclically. Then the triangles KAKBKC and XYZ are perspective, and their perspector is X3333. Contributed by Cosmin Pohoata, April 4, 2008.

(a - b + c)(a + b - c)(a + b + c) + 4abc

shows that X3333 lies on the line IO of the incenter and the circumcenter, which is also the line X1X57; viz., X57 has 1st trilinear (a - b + c)(a + b - c) and X1 has 1st trilinear 1, so that the above representation may be viewed as a linear combination using as coefficients the symmetric functions a + b + c and 4abc. The representation generalizes to

(a - b + c)(a + b - c)(a + b + c) + kabc,

where k is a symmetric function of degree 0 in a,b,c. Every point on the line IO necessary has such a form. (Indeed, every point on every line is given by such linear combinations of any two points on the line.) For the line IO, Peter Moses contributed (April 24, 2008) a list of points with corresponding functions k(a,b,c). Aside from inradius r and circumradius R, other symbols used are identified elsewhere in ETC as indicated, or else are formulated here: s = (a+b+c)/2; J as at X1113; e as at X1340; \(S_{A}\), \(S_{B}\) \(S_{C}\) as at X1117; and \(s_{A}\) = (-a+b+c)/2, \(s_{B}\) = (a-b+c), \(s_{C}\) = (a+b-c)/2.

X k 1 (k infinite) 3 k = -2(r + 2R)/R 35 k = -2(r + 3R)/R 36 k = -2(r + R)/R 40 k = -4 46 k = -2 55 k = -2(r + 4R)/R 56 k = -2r/R 57 k = 0 65 k = 2r/R 165 k = -(r +4r)/R 171 k = -(\(r^{2}\) + 4rR - \(s^{2})\)/(rR) 241 k = -\(2s^{2}\)/(rR + \(4R^{2})\) 354 k = 2(r + 4R)//R 484 k = -3 559 k = 2(\(3^{1/2}\)s//R 940 k = (a + b + c)3/(abc) 942 k = 2(r + 2R)/R 980 k = -\(2s^{2}(r^{2} + s^{2})\)/(\(rR(r^{2}\) + 4rR + \(s^{2})\)) 982 k = -(a + b + c)(\(a^{2}\) + \(b^{2}\) + \(c^{2})\)/(abc) 986 k = (\(r^{2}\) - \(s^{2})\)/(rR) 988 k = -(\(r^{2}\) + \(s^{2})\)/(rR) 999 k = -2(r - 2rR)/R 1038 k = (\(J^{2}\) - 4r/R - 9)/2 1040 k = - (\(J^{2}\) + 4r/R + 7)/2 1060 k = \(J^{2}\) - 2r/R - 5 1062 k = - (\(J^{2}\) + 2r/R + 3) 1082 k = -2(\(3^{1/2})\)s/R 1115 k = -2(r + 4R)/(3R) 1159 k = 2(5r + 2R)/(3R) 1214 k = -\(2s^{2}\)/(rR + \(2R^{2})\) 1319 k = -6r/R 1381 k = 2(-r - R + (\(R^{2}\) - 2rR)1/2)/R 1382 k = 2(-r - R + -(\(R^{2}\) - 2rR)1/2)/R 1385 k = -2(3r + 2R)/R 1388 k = -10r/R 1402 k = -2\(r(r^{2}\) + 4rR + \(s^{2})\)/(\(R(r^{2}\) + \(s^{2})\)) 1403 k = -2\(r(r^{2}\) + 4rR + \(s^{2})\)/(\(R(r^{2}\) - 4rR + \(s^{2})\)) 1420 k = -4r/R 1429 k = -2r(a + b + c)2/(\(R(a^{2}\) + \(b^{2}\) + \(c^{2})\)) 1454 k = -4r/(2r + R) 1460 k = 2(area)(\(a^{2}\) + \(b^{2}\) + \(c^{2})\)/(2\(R_{s}(r^{2}\) - \(s^{2})\)) 1466 k = -2r(r + 2R)/(rR - \(2R^{2})\) 1467 k = -r(r + 2R)/\(R^{2}\) 1470 k = 2r(r + R)/(\(R^{2}\) - rR) 1482 k = 2(3r - 2R)/R 1617 k = -2r(r + 4R)/(\(2R^{2}\) + rR) 1697 k = -8 1735 k = -2 + (\(J^{2}\) - 1)R/(2r) 1754 k = -(4r + (7 + \(J^{2})\)R)/(2r + \(4R^{2})\) 1758 k = -(\(r^{2}\) + 4rR + \(s^{2})\)/(2rR + \(R^{2})\) 1764 k = -\(4s^{2}\)/(\(r^{2}\) + \(s^{2})\) 1771 k = -2 + (1 - \(J^{2})\)R/(2r) 1936 k = [\(s^{2}\) - (r + 2R)(4 + 4R)]/(Rr + \(R^{2})\) 2077 k = - 4 + \(2r^{2}\) /(\(R^{2}\) - rR) 2078 k = -2(\(r^{2}\) + 4rR)/(rR + \(R^{2})\) 2093 k = (r - 2R)/R 2095 k = (\(2r^{2}\) - \(8R^{2})\)/(3rR + \(2R^{2})\) 2098 k = 2(3r - 4R)/R 2099 k = 6r/R 2446 k = -4(1 + (1 - 2r/R)1/2) + 2r/R 2447 k = -4(1 - (1 - 2r/R)1/2) + 2r/R 2448 k = -3 + (1 + 2r/R)1/2 2449 k = -3 - (1 + 2r/R)1/2 2556 k = -4 + 2r(e + E)/(Re - RE), where E = (1 - 2r/R)1/2 2557 k = -4 + 2r(e - E)/(Re + RE), where E = (1 - 2r/R)1/2 2464 k = -4(1 + e) - 2r/R 2465 k = -4(1 - e) - 2r/R 2572 k = -4 - 2r/(R + eR) 2573 k = -4 - 2r/(R - eR) 2646 k = -2(3r + 4R)/R 3057 k = 2(r - 4R)/R 3072 k = (R - \(RJ^{2}\) -4r)/(2r + 2R) 3075 k = (\(S_{A}\)\(S_{B}\)\(S_{C}\)2)/(\(abcs_{A}\)\(s_{B}\)\(s_{C}\)2) 3245 k = -2(5R - r)/(3R) 3256 k = 2r(r + 4R)/(R - \(rR^{2})\) 3303 k = -2(r + 8R)/R 3304 k = -2(r - 4R)/R 3333 k = 4 3336 k = -1 3337 k = 1 3338 k = 2 3339 k = r/R 3340 k = 4r/R 3361 k = -r/R

Let A' be the midpoint of X1 and A-intouch point. Define B' and C' cyclically. Traingle A'B'C' is homothetic to the excentral triangle, and the center of homothety is X3333. (Randy Hutson, December 2, 2017)