If the three green triangles ABF , BCD and CAE are similar, then... Have the triangles ABC and DEF the same barycenter G ?
Pepe Chapuzas imagined that these real triangles were imaginary, so he could work with complex numbers...
Dear Teacher:
Let's consider the complex numbers a, b, c, d, e, f, g associated with the points A, B, C, D, E, F, G... Since the green triangles are similar (same angles, proportional sides), then...
(d−b) / (c−b) = (e−c) / (a−c) = (f−a) / (b−a) = h
d = ch − bh + b
e = ah − ch + c
f = bh − ah + a
Hence
(d+e+f) / 3 = (a+b+c) / 3 = g
Another problem:
In a circle of radius r we have three equidistant chords (red), and the distance between any two chords is r too, then... Are the midpoints of the chords the vertices of an equilateral triangle?
Pepe Chapuzas solved it too...
Dear Teacher:
I have a circle... I can assume that its center is 0 and its radius is 2 to simplify, can I not? And I can assume that the endpoints of the chords are... (in polar coordinates):
2 α and 2 β−60°
2 β and 2 γ−60°
2 γ and 2 α−60°
The midpoints are...
a = 1 α + 1 β−60° = 1 α + 1 β · 1 300°
b = 1 β + 1 γ−60° = 1 β + 1 γ · 1 300°
c = 1 γ + 1 α−60° = 1 γ + 1 α · 1 300°
Let me calculate...
(a−b) / (c−b) =
= (1 α + 1 β · 1 300° − 1 β − 1 γ · 1 300°) / (1 γ + 1 α · 1 300° − 1 β − 1 γ · 1 300°) =
= (1 α + 1 β · 1 240° + 1 γ · 1 120°) / (1 α · 1 300° + 1 β · 1 180° + 1 γ · 1 60°) =
= 1 60°
= (1 α + 1 β · 1 300° − 1 β − 1 γ · 1 300°) / (1 γ + 1 α · 1 300° − 1 β − 1 γ · 1 300°) =
= (1 α + 1 β · 1 240° + 1 γ · 1 120°) / (1 α · 1 300° + 1 β · 1 180° + 1 γ · 1 60°) =
= 1 60°
If A , B , C are the corresponding real points to a , b , c , then the sides BA and BC have the same length and the angle between them measures 60°, that is, the triangle ABC is equilateral.
No hay comentarios:
Publicar un comentario