
Isogonal conjugate wrt cevian and pedal triangles
Dear geometers Let ABC be a triangle with orthocenter H. I see that 1. Isogonal conjugate points of P wrt cevian triangles of P wrt triangles HBC, HCA, HAB, ABC are collinear points. 2. Isogonal conju
Dear geometers Let ABC be a triangle with orthocenter H. I see that 1. Isogonal conjugate points of P wrt cevian triangles of P wrt triangles HBC, HCA, HAB, ABC are collinear points. 2. Isogonal conju

By Tran Quang Hung
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NPC center lies on Euler line 3 messages
Dear geometers, Let ABC be a triangle. NPC center N. A'B'C' is circumcevian triangle of N. Oa, Ob, Oc are circumcenter of triangle NB'C', NC'A', NA'B' respectively. Then NPC centers of triangle OaObOc
Dear geometers, Let ABC be a triangle. NPC center N. A'B'C' is circumcevian triangle of N. Oa, Ob, Oc are circumcenter of triangle NB'C', NC'A', NA'B' respectively. Then NPC centers of triangle OaObOc

By Tran Quang Hung
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Forum geom 2 messages
Dear geometers, Is forum geometrucum still active? There has been no update for almost a year. Regards, Rona
Dear geometers, Is forum geometrucum still active? There has been no update for almost a year. Regards, Rona

By rona levi
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ABOUT MALFATTI CIRCLES
Dear geometer, I have just found an inequality for the malfatti circles as follows: TXM697: Let 𝑝 be the perimeter of the Malfatti triangle of ∆ABC. Then we have: 𝑝 ≥ 𝐼𝐴 + 𝐼𝐵 + 𝐼𝐶 + 3𝑟 − 𝑠 ≥ (𝑎𝑏 + 𝑏𝑐
Dear geometer, I have just found an inequality for the malfatti circles as follows: TXM697: Let 𝑝 be the perimeter of the Malfatti triangle of ∆ABC. Then we have: 𝑝 ≥ 𝐼𝐴 + 𝐼𝐵 + 𝐼𝐶 + 3𝑟 − 𝑠 ≥ (𝑎𝑏 + 𝑏𝑐

By Minh Trịnh Xuân
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MUTUAL Polar Conic 4 messages
Dear friendsgeometers !!! Hello from RUSSIA ! Help me , please !!! How construct " MUTUAL Polar conic of ABC and Any triangle perspective and Not homothetic to ABC at W " ( See X32559 ... X32588 in E
Dear friendsgeometers !!! Hello from RUSSIA ! Help me , please !!! How construct " MUTUAL Polar conic of ABC and Any triangle perspective and Not homothetic to ABC at W " ( See X32559 ... X32588 in E

By kitaisky viktor
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ABOUT ARBITRARY CIRCLES 4 messages
TXM697: Let us assume that two circles (S) and (S’) are tangent to the three given circles (A), (B), (C) in either of the following two cases (Please see attached photo): Then we have: SA + S’A = SB +
TXM697: Let us assume that two circles (S) and (S’) are tangent to the three given circles (A), (B), (C) in either of the following two cases (Please see attached photo): Then we have: SA + S’A = SB +

By Minh Trịnh Xuân
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ABOUT ARBITRARY CIRCLE
Dear geometer It took me more than 5 pages to get this formula. TXMR: Let C₀(O, R) be the radical circle of the three given circles C₁(A, r₁), C₂(B, r₂), C₃(C, r₃). Let us assume that two circles Q₁(
Dear geometer It took me more than 5 pages to get this formula. TXMR: Let C₀(O, R) be the radical circle of the three given circles C₁(A, r₁), C₂(B, r₂), C₃(C, r₃). Let us assume that two circles Q₁(

By Minh Trịnh Xuân
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Six Isodynamic Points on a Circle 2 messages
Dear Geometers, Let K=X(6)Symmedian point of ABC DEF, circumcevian triangle of ABC Sa: X(15) of ABD, Saa X(15) of ADC. Define Sb, Sbb, Sc, Scc cyclically. *Sa,Saa,Sb,Sbb,Sc,Scc lie on same circle. Be
Dear Geometers, Let K=X(6)Symmedian point of ABC DEF, circumcevian triangle of ABC Sa: X(15) of ABD, Saa X(15) of ADC. Define Sb, Sbb, Sc, Scc cyclically. *Sa,Saa,Sb,Sbb,Sc,Scc lie on same circle. Be

By Kadir Altıntaş
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Circles Passing through X(11)Feuerbach Point of ABC
Dear Geometers, Let X7=Gergonne point of ABC. Parallel from X7 to BC intersects the AB at B1 and AC at C1. Parallel from X7 to AC intersects the AB at A1 and BC at C2. Parallel from X7 to AB intersect
Dear Geometers, Let X7=Gergonne point of ABC. Parallel from X7 to BC intersects the AB at B1 and AC at C1. Parallel from X7 to AC intersects the AB at A1 and BC at C2. Parallel from X7 to AB intersect

By Kadir Altıntaş
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The triangle center (S^2+SB SC)/(SA(3S^2+5 SB SC))
The barycentric coordinates of X(16835) are ((S^2SB SC)/(SA(3S^25 SB SC)) : :) See Antreas Hatzipolakis and César Lozada, Hyacinthos 27397. By changing the sign  for +, the triangle center W = (S^2
The barycentric coordinates of X(16835) are ((S^2SB SC)/(SA(3S^25 SB SC)) : :) See Antreas Hatzipolakis and César Lozada, Hyacinthos 27397. By changing the sign  for +, the triangle center W = (S^2

By Angel Montesdeoca
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Euler line is OI line
Dear geometers, Let ABC be a triangle with circumcircle (O). Parallel line through to BC meets (O) at A' Define similarly the points B' and C'. Then OI line of triangle A'B'C' is Euler line of ABC. Wh
Dear geometers, Let ABC be a triangle with circumcircle (O). Parallel line through to BC meets (O) at A' Define similarly the points B' and C'. Then OI line of triangle A'B'C' is Euler line of ABC. Wh

By Tran Quang Hung
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Equilateral From Triangulation 3 messages
Dear Geometers, Let P=X(15), or X(16) of ABC. Oa, Circumcenter of PBC. Define Ob,Oc, cyclically. A'=X(125) of PObOc. Define B',C' cyclically. *A'B'C' is equilateral. **For P=X(15), center of A'B'C' ha
Dear Geometers, Let P=X(15), or X(16) of ABC. Oa, Circumcenter of PBC. Define Ob,Oc, cyclically. A'=X(125) of PObOc. Define B',C' cyclically. *A'B'C' is equilateral. **For P=X(15), center of A'B'C' ha

By Kadir Altıntaş
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A Computation for r2018 (http://www.xtec.cat/~qcastell/ttw/ttweng/resultats/r2018.html) 4 messages
Dear Geometers, Let A', B', C' be the altitude feet; Ab, Ac the reflections of A' on BB', CC'; Oa the reflection of the circumcenter O on the side BC; Ga the centroid of OaAbAc. Define analogously Gb,
Dear Geometers, Let A', B', C' be the altitude feet; Ab, Ac the reflections of A' on BB', CC'; Oa the reflection of the circumcenter O on the side BC; Ga the centroid of OaAbAc. Define analogously Gb,

By Kadir Altıntaş
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Antipedal triangle and Circumcevian triangle are perspective 4 messages
Dear all, Let A'B'C', A*B*C* be respectively the antipendal and circumcevian triangle of a point P w.r.t triangle ABC. Then triangles A'B'C', A*B*C* are perspective. i.e A'A*, B'B*, C'C* are concurren
Dear all, Let A'B'C', A*B*C* be respectively the antipendal and circumcevian triangle of a point P w.r.t triangle ABC. Then triangles A'B'C', A*B*C* are perspective. i.e A'A*, B'B*, C'C* are concurren

By Vu Thanh Tung
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ABOUT ARBITRARY CIRCLES 2 messages
Dear geometer It took me more than 5 pages to get this formula. TXMR: Let C₀(O, R) be the radical circle of the three given circles C₁(A, r₁), C₂(B, r₂), C₃(C, r₃). Let us assume that two circles Q₁(
Dear geometer It took me more than 5 pages to get this formula. TXMR: Let C₀(O, R) be the radical circle of the three given circles C₁(A, r₁), C₂(B, r₂), C₃(C, r₃). Let us assume that two circles Q₁(

By Minh Trịnh Xuân
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Circlecevian, Triangle Centers, Cyclologic, Perspective 12 messages
Dear all, In the message 11, the notation of Circlecevian Triangle is mentioned. We found some interesting properties: Let Pa Pb Pc be the circlecevian triangle of a point P w.r.t. a triangle ABC. Let
Dear all, In the message 11, the notation of Circlecevian Triangle is mentioned. We found some interesting properties: Let Pa Pb Pc be the circlecevian triangle of a point P w.r.t. a triangle ABC. Let

By Vu Thanh Tung
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Rquilateral Triangle from X(125)'s of Subtriangles
Dear Geometers, Let P=X(13) or X(14) of ABC. A'=X(125) of PBC. Define B',C' cyclically. *A'B'C' are equilateral triangles. Interestingly these equilateral triangles have same centers with equilaretal
Dear Geometers, Let P=X(13) or X(14) of ABC. A'=X(125) of PBC. Define B',C' cyclically. *A'B'C' are equilateral triangles. Interestingly these equilateral triangles have same centers with equilaretal

By Kadir Altıntaş
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An instance of the nine point circle as radical circle, or and appearance of X3613. 2 messages
Given an acute triangle ABC with CA > AB and X a point on BC, call Xb the intersection of CA and the perpendicular to BC at X. If XbXC halves the area of ABC, prove that the circle (C, CX) is orthogon
Given an acute triangle ABC with CA > AB and X a point on BC, call Xb the intersection of CA and the perpendicular to BC at X. If XbXC halves the area of ABC, prove that the circle (C, CX) is orthogon

By Francisco Javier García Capitán
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Parabola locus of centres of hyperbolas 2 messages
The polars of the points of a line d, with respect to the circumcircle and the Steiner circumellipse of a triangle, intersect at points on a hyperbola H(d). When the line d rotates around a point U, t
The polars of the points of a line d, with respect to the circumcircle and the Steiner circumellipse of a triangle, intersect at points on a hyperbola H(d). When the line d rotates around a point U, t

By Angel Montesdeoca
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Perspective and Parallelogic
Dear all, The Sondat theorem states that if two triangles are both orthologic and perspective, then the perspector and two orthologic centers are collinear. Further, the line containing these three po
Dear all, The Sondat theorem states that if two triangles are both orthologic and perspective, then the perspector and two orthologic centers are collinear. Further, the line containing these three po

By Vu Thanh Tung
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