Τρίτη 30 Απριλίου 2013

EQUILATERAL TRIANGLE

Lemma:

Let 123456 be a hexagon insrcribed in circle (0).

If the triangles 012, 034, 056 are equilateral, then the triangle with vertices the midpoints of the segments 23, 45, 61 is equilateral.

Reference:

Mathematical Chronicles (Athens), #5 (May 1969), p. 87

APH, Anopolis, #197

Application:

Let ABC be a triangle. Construct the equilateral triangle OAbAc with OA as altitude from O to base AbAc (Ab, Ac near to B, C, resp. Similarly construct the equilateral triangles OBaBc, OCaCb.

Let Ma, Mb, Mc be the midpoints of BcCb, CaAc, AbBa, resp.

The triangle MaMbMc is equilateral with center the centroid of ABC

The Ma, Mb, Mc are the centers of the equilateral A'BC, B'CA, C'AB, erected outwardly on the sides of the triangle.

The triangles ABC, MaMbMc are perspective.

The perspector is 1st Napoleon point.

Antreas P. Hatzipolakis, 30 April 2013

Δεν υπάρχουν σχόλια:

Δημοσίευση σχολίου

Cosmology of Plane Geometry: Concepts and Theorems

Alexander Skutin,Tran Quang Hung, Antreas Hatzipolakis, Kadir Altintas: Cosmology of Plane Geometry: Concepts and Theorems> ΨΗΦ. C...