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Medialtial

(4)
[APH] Hyacinthos 253 Definition: Mi : the midpoints of sides BC,CA,AB, resp. Hi :
Antreas Hatzipolakis
4
posts
11:50 AM

Rectangles erected on sides of a triangle

(3)
[APH]: Let ABC be a triangle. Let BCCaBa, CAAbCb, ABBcAc be three arbitrary rectangles erected on the sides BC, CA, AB, resp. outwardly ABC. The perpendicular
Antreas Hatzipolakis
3
posts
Sep 22

Collinear points, Coaxial circles

(3)
[APH]: Let ABC be a triangle, P a point and A1B1C1 the cevian triangle of P. Let K be the point of concurrence of the circumcircles of AB1C1, BC1A1, CA1B1,
Antreas Hatzipolakis
3
posts
Sep 20

Perspective Triangles

(3)
[Stalislav Takhayev]: Any triangle of ABC, I-incenter. The outer Soddy circle is constructed.Points of contact of three mutually tangent of a circles with
Antreas Hatzipolakis
3
posts
Sep 17

Bicentric points

(4)
[APH] Let ABC be a triangle. The perpendicular to AB at A intersects BC at A1 The perpendicular to BC at B intersects CA at B1 The perpendicular to CA at C
Antreas Hatzipolakis
4
posts
Sep 12

I, Reflected Euler lines

(7)
[APH]: Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P. Denote: La, Lb, Lc = the Euler lines of PBC, PCA, PAB, resp. L1, L2, L3 = the
Antreas Hatzipolakis
7
posts
Sep 12

Orthogonal circle

(3)
[APH]: Let ABC be a triangle and A'B'C', A"B"C" the pedal triangles of O,H, resp. The orthogonal circle ( = radical circle) of the circles (A', A'A"), (B',
Antreas Hatzipolakis
3
posts
Sep 12

N, Reflected Euler lines

(5)
[APH]: Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P. Denote: La, Lb, Lc = the Euler lines of AB'C', BC'A', CA'B', resp. L1, L2, L3 = the
Antreas Hatzipolakis
5
posts
Sep 9

A problem related to Artzt parabolas

(3)
[APH]: A Greek mathematician (F. D. Constantinides) published a paper on Artzt parabolas, where writes the following conjecture: If from a given triangle ABC
Antreas Hatzipolakis
3
posts
Sep 3
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