A 3-point of a triangle is defined as a point where there exists
three distinct lines through the point and each line divides the
triangle into two polygons of equal area. Clearly, the centroid of the
triangle is a 3-point of the triangle. What percent of the area of a
triangle is the area of the set of 3-points of the triangle?
gdb
Currently, 1114 triangle centers are known.
http://cedar.evansville.edu/%7Eck6/encyclopedia/
--Ed Pegg Jr.
That problem it isn´t very concerned by the Encyclopedia of Triangle
Centers ...
It is known that the tangents to a hyperbola determine on the
asymptotes segments whose product is constant, and that the point of
tangencia T is the midpoint of the segment that the asymptotes
determine in the tangent.
Therefore the segments with ends in the sides to and b, that bisect to
the triangle, are tangent in their midpoints to a hyperbola that it
has to the sides to and b like asymptotes.
Calling X, Y and Z to medium the midpoints of the medians m_a, m_b and
m_c, such hyperbola is tangent to m_a in X and m_b in and.
The problem has 3 solutions if point R is inside tricuspoid XYZ
composed by three arcs of hyperbola that they have to each pair of
sides like asymptotes, 2 solutions on the own tricuspoid, excepted
points X, and and Z, and 1 in all the rest of the plane.
That is an affine problem, the ratio of the areas is invariant under
affine transformations. Therefore, it can be solved in any triangle
with full generallity. In an equilateral triangle, by example, or in
the triangle with vertices A(0, 0), B(1, 0) and C(0, 1). In any case,
it is easy, by
integration, to calculate the area of that tricuspoid.
Its area small is compared with the one of the triangle, exactly
(3*ln(2)-2)/4=0.01986... of the one of the triangle, less of a 2%.
P.S.: This is an automatic translate from Spanish. Let be indulgent
with the grammar ...
--
Saludos,
Ignacio Larrosa Cañestro
A Coruña (España)
ignacio...@eresmas.net
ICQ #94732648
Hi Ignacio,
I agree with your response in everyway ( and the translation was good
enough to follow ). Thanks for the name of the set - tricuspoid - I searched
the internet and found many interesting sites. Some of the sites with JAVA
applets are fantastic.
Since the ratio of the areas is invariant, I used an equalateral triangle
ABC of side length one. Let <ST> denote the vector from point S to point T.
For every point in the plane of triangle ABC, <AP> = u <AB> + v <AC>
for some real numbers u and v. By symmetry I decided to integrate the area
between median m_a [ v = u ], median m_b [ v = (1-u)/2 ], and the arc of the
tricuspoid from your point Y to your point Z [ v = 1/(8u) ]. This is 1/6 th
of the area tricuspoid. For the limits of integration I used the points that
satisfied these three equations pairwise : the centroid (1/3,1/3), your
point Y (1/2,1/4), and the midpoint of the arc
(sqrt(2)/4,sqrt(2)/4). All I needed to know was how to integrate (u^n)du for
n = -1,0, and 1. Also, I had to integrate (1-u)du from 0 to 1 to find the
area of triangle ABC ( =1/2 ).
It would have been nice if the points X, Y, and Z were "2-points", but no
such luck.
gdb
I used the name 'tricuspoid' in a genericall sense, i.e. 'curve with
three cusp'. That name usually stand for the Deltoid of Steiner of any
triangle, that is an hypocicloid of three cusps, but it haven't relatión
with the recint we are talking. This is the curve you probably has found
in many Internet pages. It is the envelope of the Simson's lines of the
triangle.
The border of the 3-points set of a triangle, with a 'similar' (in a
generic sense) shape that the Steiner's Deltoid, is composite of three
hyperbolic arcs.
Other important difference is that the shape of the Steiner's Deltoid is
unique for any triangle, while the border of the 3-points set no.
--
Saludos,
Ignacio Larrosa Cañestro
A Coruña (España)
ignacio...@eresmas.net
ICQ #94732648
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