X3548
EULER LINE INTERCEPT OF LINE X68 X_125
5, 6, 7
3, 4, 5
Kepler
Kimberling
< A B C >: Reference Triangle △ABC
< M_a M_b M_c >: Medial Triangle △M_a M_b M_c
[ A B ]: Sideline AB (Base c = 1)
[ B C ]: Sideline BC (Side a)
[ C A ]: Sideline CA (Side b)
{ 2, 3542 }: Euler Line
{ 2, 3 }: Euler Line
{ 68, 125 }: Central Line L(68, 125)
{ 498, 1060 }: Central Line L(498, 1060)
{ 499, 1062 }: Central Line L(499, 1062)
{ 1204, 1568 }: Central Line L(1204, 1568)
A = Vertex A (0.0000, 0.0000)
B = Vertex B (1.0000, 0.0000)
C = Vertex C (0.6122, 0.5999)
M_a = Midpoint M_a of BC (0.8061, 0.2999)
M_b = Midpoint M_b of CA (0.3061, 0.2999)
M_c = Midpoint M_c of AB (0.5000, 0.0000)
X_2 = CENTROID (0.5374, 0.2000)
X_3542 = EULER LINE INTERCEPT OF LINE X(33)X(498) (0.6774, 0.5661)
X_3 = CIRCUMCENTER (0.5000, 0.1021)
X_68 = PRASOLOV POINT (-0.0535, -0.0055)
X_125 = CENTER OF JERABEK HYPERBOLA (0.6803, 0.0260)
X_498 = YFF CONCURRENT CONGRUENT CIRCLES POINT (0.5518, 0.2140)
X_1060 = INTERSECTION OF LINES X(1)X(3) AND X(5)X(34) (0.5181, 0.1353)
X_499 = {X(1),X(2)}-HARMONIC CONJUGATE OF X(498) (0.4458, 0.1102)
X_1062 = INTERSECTION OF LINES X(1)X(3) AND X(5)X(33) (0.4633, 0.0347)
X_1204 = 3rd SARAGOSSA POINT OF X(64) (0.6497, 0.1923)
X_1568 = ORTHOJOIN OF X(800) (0.1362, -0.3019)
X_3548 = EULER LINE INTERCEPT OF LINE X(68)X(125) (0.4674, 0.0169)
A
B
C
Ma
Mb
Mc
X2
X3542
X3
X68
X125
X498
X1060
X499
X1062
X1204
X1568
X3548
(a, b, c) = (5, 6, 7) | AB = 1
< A B C >: Reference Triangle △ABC
< M_a M_b M_c >: Medial Triangle △M_a M_b M_c
[ A B ]: Sideline AB (Base c = 1)
[ B C ]: Sideline BC (Side a)
[ C A ]: Sideline CA (Side b)
{ 2, 3542 }: Euler Line
{ 2, 3 }: Euler Line
{ 68, 125 }: Central Line L(68, 125)
{ 498, 1060 }: Central Line L(498, 1060)
{ 499, 1062 }: Central Line L(499, 1062)
{ 1204, 1568 }: Central Line L(1204, 1568)
A = Vertex A (0.0000, 0.0000)
B = Vertex B (1.0000, 0.0000)
C = Vertex C (0.6400, 0.4800)
M_a = Midpoint M_a of BC (0.8200, 0.2400)
M_b = Midpoint M_b of CA (0.3200, 0.2400)
M_c = Midpoint M_c of AB (0.5000, 0.0000)
X_2 = CENTROID (0.5467, 0.1600)
X_3542 = EULER LINE INTERCEPT OF LINE X(33)X(498) (0.6400, 0.4800)
X_3 = CIRCUMCENTER (0.5000, 0.0000)
X_68 = PRASOLOV POINT (-0.0283, 0.1600)
X_125 = CENTER OF JERABEK HYPERBOLA (0.6400, 0.0000)
X_498 = YFF CONCURRENT CONGRUENT CIRCLES POINT (0.5709, 0.1782)
X_1060 = INTERSECTION OF LINES X(1)X(3) AND X(5)X(34) (0.5000, 0.0000)
X_499 = {X(1),X(2)}-HARMONIC CONJUGATE OF X(498) (0.2800, -0.0400)
X_1062 = INTERSECTION OF LINES X(1)X(3) AND X(5)X(33) (0.5000, 0.0000)
X_1204 = 3rd SARAGOSSA POINT OF X(64) (0.6400, 0.0000)
X_1568 = ORTHOJOIN OF X(800) (0.3600, 0.0000)
X_3548 = EULER LINE INTERCEPT OF LINE X(68)X(125) (0.5000, 0.0000)
A
B
C
Ma
Mb
Mc
X2
X3542
X3
X68
X125
X498
X1060
X499
X1062
X1204
X1568
X3548
(a, b, c) = (3, 4, 5) | AB = 1
< A B C >: Reference Triangle △ABC
< M_a M_b M_c >: Medial Triangle △M_a M_b M_c
[ A B ]: Sideline AB (Base c = 1)
[ B C ]: Sideline BC (Side a)
[ C A ]: Sideline CA (Side b)
{ 2, 3542 }: Euler Line
{ 2, 3 }: Euler Line
{ 68, 125 }: Central Line L(68, 125)
{ 498, 1060 }: Central Line L(498, 1060)
{ 499, 1062 }: Central Line L(499, 1062)
{ 1204, 1568 }: Central Line L(1204, 1568)
A = Vertex A (0.0000, 0.0000)
B = Vertex B (1.0000, 0.0000)
C = Vertex C (0.6180, 0.4859)
M_a = Midpoint M_a of BC (0.8090, 0.2429)
M_b = Midpoint M_b of CA (0.3090, 0.2429)
M_c = Midpoint M_c of AB (0.5000, 0.0000)
X_2 = CENTROID (0.5393, 0.1620)
X_3542 = EULER LINE INTERCEPT OF LINE X(33)X(498) (0.6180, 0.4859)
X_3 = CIRCUMCENTER (0.5000, 0.0000)
X_68 = PRASOLOV POINT (-0.0357, 0.1620)
X_125 = CENTER OF JERABEK HYPERBOLA (0.6180, 0.0000)
X_498 = YFF CONCURRENT CONGRUENT CIRCLES POINT (0.5596, 0.1801)
X_1060 = INTERSECTION OF LINES X(1)X(3) AND X(5)X(34) (0.5000, 0.0000)
X_499 = {X(1),X(2)}-HARMONIC CONJUGATE OF X(498) (0.3290, -0.0269)
X_1062 = INTERSECTION OF LINES X(1)X(3) AND X(5)X(33) (0.5000, 0.0000)
X_1204 = 3rd SARAGOSSA POINT OF X(64) (0.6180, 0.0000)
X_1568 = ORTHOJOIN OF X(800) (0.3820, 0.0000)
X_3548 = EULER LINE INTERCEPT OF LINE X(68)X(125) (0.5000, 0.0000)
A
B
C
Ma
Mb
Mc
X2
X3542
X3
X68
X125
X498
X1060
X499
X1062
X1204
X1568
X3548
Kepler (1, √φ, φ) | AB = 1
< A B C >: Reference Triangle △ABC
< M_a M_b M_c >: Medial Triangle △M_a M_b M_c
[ A B ]: Sideline AB (Base c = 1)
[ B C ]: Sideline BC (Side a)
[ C A ]: Sideline CA (Side b)
{ 2, 3542 }: Euler Line
{ 2, 3 }: Euler Line
{ 68, 125 }: Central Line L(68, 125)
{ 498, 1060 }: Central Line L(498, 1060)
{ 499, 1062 }: Central Line L(499, 1062)
{ 1204, 1568 }: Central Line L(1204, 1568)
A = Vertex A (0.0000, 0.0000)
B = Vertex B (1.0000, 0.0000)
C = Vertex C (0.6331, 0.2801)
M_a = Midpoint M_a of BC (0.8166, 0.1400)
M_b = Midpoint M_b of CA (0.3166, 0.1400)
M_c = Midpoint M_c of AB (0.5000, 0.0000)
X_2 = CENTROID (0.5444, 0.0934)
X_3542 = EULER LINE INTERCEPT OF LINE X(33)X(498) (0.5730, 0.3308)
X_3 = CIRCUMCENTER (0.5000, -0.2747)
X_68 = PRASOLOV POINT (0.7351, 0.0696)
X_125 = CENTER OF JERABEK HYPERBOLA (0.5678, -0.0079)
X_498 = YFF CONCURRENT CONGRUENT CIRCLES POINT (0.5865, 0.1151)
X_1060 = INTERSECTION OF LINES X(1)X(3) AND X(5)X(34) (0.6711, 0.3254)
X_499 = {X(1),X(2)}-HARMONIC CONJUGATE OF X(498) (0.7689, 0.2093)
X_1062 = INTERSECTION OF LINES X(1)X(3) AND X(5)X(33) (0.5870, 0.0306)
X_1204 = 3rd SARAGOSSA POINT OF X(64) (0.5298, -0.4963)
X_1568 = ORTHOJOIN OF X(800) (0.5303, 0.2721)
X_3548 = EULER LINE INTERCEPT OF LINE X(68)X(125) (0.5301, -0.0254)
A
B
C
Ma
Mb
Mc
X2
X3542
X3
X68
X125
X498
X1060
X499
X1062
X1204
X1568
X3548
Kimberling (6, 9, 13) | AB = 1
X3548 (EULER LINE INTERCEPT OF LINE X68 X_125 ) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\) .
Trilinears: \(\sec B \sec C - 4 \sin B \sin C : :\)
Search Value (6, 9, 13): \(3.969030629239\)
▸ Incident Central Lines7 lines
{2 , 3 }, {68 , 125 }, {254 , 2970 }, {498 , 1060 }, {499 , 1062 }, {1147 , 1899 }, {1204 , 1568 }
▸ Functional Relations1 relation
Notes
As a point on the Euler line, X3548 has Shinagawa coefficients \((E - 4F, -E)\) .
Let La be the polar of X4 wrt the circle centered at A and passing through X3 , and define Lb and Lc cyclically. (Note that X4 is the perspector of any circle centered at a vertex of ABC .) Let A″ = Lb ∩Lc , and define B″ and C″ cyclically. Triangle A″B″C″ is homothetic to the medial triangle , and the center of homothety is X3548 . (Randy Hutson , December 11, 2015)