1st Power Point
Antreas P. Hatzipolakis
These are (as they are indexed in p. 293):
X_31: 2nd Power Point. Trilinears: (a^2::)
X_32: 3rd Power Point: Trilinears: (a^3::)
X_75: Unnamed Trilinears: (a^(-2)::)
X_76: 3rd Brocard Point: Trilinears: (a^(-3)::)
X_365: Square Root Point: Trilinears: (a^(1/2)::)
X_366: Isogonal Conjuagate of X_365: Trilinears: (a^(-1/2)::)
Now, the question is: Which is the 1st Power Point?
This point should be the point with trilinears (a^1::).
Which is the geometrical importance of this point?
Well... it is the triangle point with the property: x^2 + y^2 + y^2 = min,
where x, y, z are the distances from the point to BC,CA,AB, respectively
Note 1: I looked at the _TCCT_ Index, and thereafter at each one point listed
there (that's the role of an Index, after all!); not at every point X_1-X_400.
So, I am not sure if the point is already among the 400s in the book.
Note 2: TCCT stands for:
Triangle Centers and Central Triangles, by Clark Kimberling.
Congressus Numerantium, Volume 129, August, 1998.
Winnipeg, Canada
[This special volume is published in order to make available the research on
Triangle Points and Central Triangles]
Students may try to prove the point's property above.
A proof can be found after some blank lines.
Antreas
There was more imagination in the head of Archimedes than in that of Homer.
-- Voltaire (1694-1778)
SPOILER
Problem:
Find a point P on a triangle ABC such that: x^2 + y^2 + z^2 = min,
where x,y,z are the distances from P to BC,CA,AB, respectively.
Lemma: Let x,y,z be Reals with ax+by+cz = d^2, where a,b,c,d fixed.
x^2 + y^2 + z^2 = min ==> x/a = y/b = z/c
Proof:
Langrange Identity: (a^2 + b^2 + c^2)*(x^2 + y^2 + z^2) - (ax + by + cz)^2 =
(ay - bx)^2 + (bz - cy)^2 + (cx - az)^2 ==>
d^2 + (ay - bx)^2 + (bz - cy)^2 + (cx - az)^2
x^2 + y^2 +z^2 = ---------------------------------------------- (1)
a^2 + b^2 + c^2
(1) = min ==> (ay - bx)^2 + (bz - cy)^2 + (cx - az)^2 = min (2)
(2) = min ==> it equals to 0 (since (ay - bx)^2, (bz - cy)^2, (cx - az)^2
are possitives) ==> (ay - bx = bz - cy = cx - az = 0) ==>
x y z
--- = --- = ---
a b c
Solution of the Problem:
A
/ \
/ \
c / \ b
/ z y \
/ P \
/ x \
B------------C
a
area(PBC) + area(PCA) + area(PAB) = area(ABC) ==>
ax + by + cz = d^2 (:area of ABC)
According to Lemma, x^2 + y^2 + z^2 = min ==> x/a = y/b = c/z
that is, the trilinears of P are (a^1::)
Note:
We learnt this problem in high school (in algebra, not in geometry)
APH
Antreas P. Hatzipolakis
>Proof:
>Langrange Identity: (a^2 + b^2 + c^2)*(x^2 + y^2 + z^2) - (ax + by + cz)^2 =
Lagrange [= La Grange]
---
>ax + by + cz = d^2 (:area of ABC)
(:twice the area of ABC)
APH
Floor van Lamoen
<...>
> Note 1: I looked at the _TCCT_ Index, and thereafter at each one point listed
> there (that's the role of an Index, after all!); not at every point X_1-X_400.
> So, I am not sure if the point is already among the 400s in the book.
The point is also known as the Symmedian point or Lemoine point... X_6
in TCCT.
Best regards,
Floor.
Antreas P. Hatzipolakis
Yes. I should have look at the previous (to 2nd Power Point) ones, since I was
not remembering that, that property is of Lemoine's!
Indeed, X_6 reads:
Symmedian Point (Lemoine Point) alpha = a, or alpha = sin A, the point
of concurrence of the symmedians (reflections of medians about corresponding
angle bisectors). X_6 is the point (alpha, beta, gamma), given here in
<i>actuasl</i> trilinear distances, that minimizes alpha^2 + beta^2 + gamma^2.
Thanks!
Antreas
Steve Sigur
>Clark Kimberling in _TCCT_ lists some so called "Power Points".
>
>These are (as they are indexed in p. 293):
>
>X_31: 2nd Power Point. Trilinears: (a^2::)
>X_32: 3rd Power Point: Trilinears: (a^3::)
>X_75: Unnamed Trilinears: (a^(-2)::)
>X_76: 3rd Brocard Point: Trilinears: (a^(-3)::)
>X_365: Square Root Point: Trilinears: (a^(1/2)::)
>X_366: Isogonal Conjuagate of X_365: Trilinears: (a^(-1/2)::)
>
>Now, the question is: Which is the 1st Power Point?
>
>This point should be the point with trilinears (a^1::).
I will use the more symmetrical conventions :b: = (:b:) = (a:b:c)
>
>Which is the geometrical importance of this point?
>Well... it is the triangle point with the property: x^2 + y^2 + y^2 = min,
>where x, y, z are the distances from the point to BC,CA,AB, respectively
Antreas,
The first power point is the symmedian point. The 3rd power point is
probably significant, being on the Brocard Line.
But these names are not good because they presume a particular coordinate
system. The first power point in trilinears is the second power point in
barycentrics.
There is an interesting story related to these points, however. A few
years ago Clark Kimberling proposed a definition of triangle centers
which implied a group law on triangle centers. If :y1: and :y2: stand
for the homogeneous coordinates of two points then the group operation is
:y1 y2:, ie, you multiply the corresponding coordinates.
The structure of this group is that of a free group with an infinite
number of generators. One of the generators is the first power point :b:
, its orbit being all the power points :b^n: for integer n.
One can plot this orbit. The positive powers all lie on a curve that goes
between the incenter and a vertex (I believe that of the largest angle).
The problem is: as far as we know, almost all of these points are
irrelevant! We know of no reason for their significance, other than that
their coordinates are powers of the sides.
So, other than :b: and :bbb:, I have not worried much about these points
(although in the back of my mind, I think :bb: has shown itself
somewhere).
Steve