X1143
2nd MALFATTI-RABINOWITZ POINT
Along each side of
ABC there is a segment that is a common tangent to two of the three Malfatti circles of
ABC . Let
A' ,
B' ,
C' be the midpoints of these respective segments. Then triangle
A'B'C' isperspective to
ABC , and the perspector is
X1143 . (
Stanley Rabinowitz , #4611, 12/29/01) For coordinates, see
Paul Yiu , #4615, 12/30/01, and
5, 6, 7
3, 4, 5
Kepler
Kimberling
< 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) { 8, 177 }: Central Line L(8, 177) { 174, 175 }: Central Line L(174, 175) { 558, 1488 }: Central Line L(558, 1488)
A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_1489 = 3rd STEVANOVIC POINT (0.5675, 0.2288) X_2 = CENTROID (0.5374, 0.2000) X_8 = NAGEL POINT (0.4694, 0.1333) X_177 = 1st MID-ARC POINT (0.5862, 0.2654) X_174 = YFF CENTER OF CONGRUENCE (0.5813, 0.2549) X_175 = ISOPERIMETRIC POINT (0.4271, -0.1006) X_558 = (X(1),X(57))-ANSWER TO QUESTION A (0.6164, 0.3210) X_1488 = 2nd STEVANOVIC POINT (0.5582, 0.2148) X_1143 = 2nd MALFATTI-RABINOWITZ POINT (0.5857, 0.2649)
A
B
C
X1489
X2
X8
X177
X174
X175
X558
X1488
X1143
(a, b, c) = (5, 6, 7) | AB = 1
< 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) { 8, 177 }: Central Line L(8, 177) { 174, 175 }: Central Line L(174, 175) { 558, 1488 }: Central Line L(558, 1488)
A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6400, 0.4800) X_1489 = 3rd STEVANOVIC POINT (0.5944, 0.1948) X_2 = CENTROID (0.5467, 0.1600) X_8 = NAGEL POINT (0.4400, 0.0800) X_177 = 1st MID-ARC POINT (0.6174, 0.2453) X_174 = YFF CENTER OF CONGRUENCE (0.6119, 0.2308) X_175 = ISOPERIMETRIC POINT (0.3600, -0.4800) X_558 = (X(1),X(57))-ANSWER TO QUESTION A (0.6515, 0.3126) X_1488 = 2nd STEVANOVIC POINT (0.5818, 0.1764) X_1143 = 2nd MALFATTI-RABINOWITZ POINT (0.6168, 0.2447)
A
B
C
X1489
X2
X8
X177
X174
X175
X558
X1488
X1143
(a, b, c) = (3, 4, 5) | AB = 1
< 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) { 8, 177 }: Central Line L(8, 177) { 174, 175 }: Central Line L(174, 175) { 558, 1488 }: Central Line L(558, 1488)
A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6180, 0.4859) X_1489 = 3rd STEVANOVIC POINT (0.5793, 0.1968) X_2 = CENTROID (0.5393, 0.1620) X_8 = NAGEL POINT (0.4499, 0.0817) X_177 = 1st MID-ARC POINT (0.5989, 0.2484) X_174 = YFF CENTER OF CONGRUENCE (0.5941, 0.2335) X_175 = ISOPERIMETRIC POINT (0.3820, -0.4859) X_558 = (X(1),X(57))-ANSWER TO QUESTION A (0.6279, 0.3174) X_1488 = 2nd STEVANOVIC POINT (0.5687, 0.1780) X_1143 = 2nd MALFATTI-RABINOWITZ POINT (0.5983, 0.2477)
A
B
C
X1489
X2
X8
X177
X174
X175
X558
X1488
X1143
Kepler (1, √φ, φ) | AB = 1
< 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) { 8, 177 }: Central Line L(8, 177) { 174, 175 }: Central Line L(174, 175) { 558, 1488 }: Central Line L(558, 1488)
A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6331, 0.2801) X_1489 = 3rd STEVANOVIC POINT (0.6100, 0.1259) X_2 = CENTROID (0.5444, 0.0934) X_8 = NAGEL POINT (0.4024, 0.0200) X_177 = 1st MID-ARC POINT (0.6266, 0.1906) X_174 = YFF CENTER OF CONGRUENCE (0.6235, 0.1737) X_175 = ISOPERIMETRIC POINT (0.7406, 0.8901) X_558 = (X(1),X(57))-ANSWER TO QUESTION A (0.6431, 0.2432) X_1488 = 2nd STEVANOVIC POINT (0.5987, 0.1042) X_1143 = 2nd MALFATTI-RABINOWITZ POINT (0.6262, 0.1903)
A
B
C
X1489
X2
X8
X177
X174
X175
X558
X1488
X1143
Kimberling (6, 9, 13) | AB = 1
X1143 (2nd MALFATTI-RABINOWITZ POINT ) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\) .
Trilinears: \(\csc A \tan A/4 : \csc B \tan B/4 : \csc C \tan C/4\)
Barycentrics: \(\tan A/4 : \tan B/4 : \tan C/4\)
Search Value (6, 9, 13): \(0.982123961231\)
▸ Alternate Coordinates1 form
Barycentrics: \(f(a,b,c) : f(b,c,a) : f(c,a,b)\) , where \(f(a,b,c) = (b + c - a)^{1/2}[2b^{1/2}c^{1/2} - (a + b + c)^{1/2}(b + c - a)^{1/2}]\)
X1143
▸ Incident Central Lines3 lines
▸ Functional Relations3 relations
Notes
Milorad R. Stevanovic, “Triangle Centers Associated with the Malfatti Circles,” :cite:`Forum Geometricorum <forum-geometricorum>` 3 (2003) 83-93.
Let A' be the touchpoint of the line BC and the incircle of the triangle BCI , where I = incenter of ABC . Define B' and C' cyclically. The line AA' , BB' , CC' concur in X1143 Randy Hutson , 9/23/2011)
