## Re: [EMHL] Re:Line

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• Dear Francisco, By the way, If a line L is tangent at P to the circumcircle of ABC and the distances of A, B, C from L are p, q, r do you know if it is
Message 1 of 19 , Jan 9, 2012
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Dear Francisco,
By the way,
If a line L is tangent at P to the circumcircle of ABC
and the distances of A, B, C from L are p, q, r
do you know if it is possible to find the points P
for which the sum p^2 + q^2 + r^2 becomes min or max?

Best regards

> Dear Bernard, Martin and Chris, I
> need to make a correction:
>
> The line for which the minimum is attached is always the
> line
> (b^2 - c^2) x + (a^2 - b^2 - T) y + (-a^2 + c^2 + T) z == 0
> and this is always the MAJOR axis of Steiner
> circum-ellipse.
>
• A remark. If the triangle is equilateral then there is no min or max because p^2 + q^2 + r^2 = 9R^2/2 and obviously p + q + r = 3R ND
Message 2 of 19 , Jan 10, 2012
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A remark.
If the triangle is equilateral then there is no
min or max because p^2 + q^2 + r^2 = 9R^2/2
and obviously p + q + r = 3R

ND

> Dear Francisco,
> By the way,
> If a line L is tangent at P to the circumcircle of ABC
> and the distances of A, B, C from L are p, q, r
> do you know if it is possible to find the points P
> for which the sum p^2 + q^2 + r^2 becomes min or max?
>
> Best regards
• Dear Francisco, Bernard, Martin and Nikos, Like Francisco stated the major Axis of the Steiner CircumEllipses is the line with the minimal sum of squared
Message 3 of 19 , Jan 10, 2012
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Dear Francisco, Bernard, Martin and Nikos,

Like Francisco stated the major Axis of the Steiner CircumEllipses is the line with the minimal sum of squared distances to the Triangle vertices. Besides that I noticed that of all lines through the Centroid the minor Axis is the line with the maximum sum of squared distances to the Triangle vertices.

New question:
When we have 4 points (1:0:0),(0:1:0),(0:0:1),(p:q:r) which line has the minimal sum of squared distances?
Does it pass through the centroid of the complete quadrangle?

Best regards,

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
>
> Dear Bernard, Martin and Chris, I need to make a correction:
>
> The line for which the minimum is attached is always the line
> (b^2 - c^2) x + (a^2 - b^2 - T) y + (-a^2 + c^2 + T) z == 0 and this is always the MAJOR axis of Steiner circum-ellipse.
>
> --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@> wrote:
> >
> > Dear Bernard, thank you.
> >
> > Yes, Coolidge, who cites Vahlen, generalizes for n points, but the minimum is attained by one single line, in our case the minor axis of the Steiner circum-ellipse.
> >
> > The minimum sum of squared distances is (1/6) (a^2 + b^2 + c^2 - 2 T) while the value for the major axis is (1/6) (a^2 + b^2 + c^2 + 2 T).
> >
> > Best regards,
> >
> > Francisco Javier.
> >
> > --- In Hyacinthos@yahoogroups.com, Bernard Gibert <bg42@> wrote:
> > >
> > > Dear Francisco,
> > >
> > > > It's a least squares problem which I believe is well known.
> > >
> > >
> > > a probable reference...
> > >
> > > J. L. Coolidge, Two geometrical applications of the method of least squares, MONTHLY 20 (1913) 187ï¿½190.
> > >
> > > Best regards
> > >
> > > Bernard
> > >
> > > [Non-text portions of this message have been removed]
> > >
> >
>
• Dear friends, ... 1. Thanks Francisco, problem 2 in Coolidge article gives the answer the minimum-square-distances-line passe through the centroid of the
Message 4 of 19 , Jan 10, 2012
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Dear friends,

> When we have 4 points (1:0:0),(0:1:0),(0:0:1),(p:q:r) which line has the minimal sum of squared distances?
> Does it pass through the centroid of the complete quadrangle?

1. Thanks Francisco, problem 2 in Coolidge article gives the answer the minimum-square-distances-line passe through the centroid of the complete quadrangle. This is also valid for all sets of n points.
2. I also found this phrase on 1st page in
http://www.mamikon.com/USArticles/SumSqrsMonthly.pdf
"Given a finite set of fixed points in 3-space, what is the locus
of a point moving in such a way that the sum of the squares of its distances from the fixed points is constant? The answer is both elegant and surprising: the locus is a sphere whose center is at the centroid of the fixed points (if we allow the empty set and a single point as degenerate cases of a sphere)."
This makes clear that for all sets of n points the sum of square distances is extending in circles (2-D) or spheres (3-D) with the centroid as center. So the Centroid has minimal sum of square distances to n points
It is not the same as square distances to a line but it gives a rough idea how square distances behave with n points.
3. In my first calculation of the minimum-square-distances-line in the 4-point case I found also 2 perpendicular lines through the Quadrangle Centroid (similar to the Steiner Axes).

Best regards,

Chris van Tienhoven
• Dear Friends, The line with minimal sum of squared distances to Triangle vertices is related to the point with minimal sum of squared distances to the sides of
Message 5 of 19 , Jan 12, 2012
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Dear Friends,

The line with minimal sum of squared distances to Triangle vertices is related to the point with minimal sum of squared distances to the sides of a triangle (duality principle).
Following the instructions from the paper of J.L. Coolidge I figured out which point in a triangle has minimal sum of squared distances to the sides of a reference triangle.
It is not the X2 (centroid), it is not X13 (Fermat Point).
It is X6 (Symmedian Point). I wouldn't have expected this.

Best regards,

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" <van10hoven@...> wrote:
>
> Dear friends,
>
> > When we have 4 points (1:0:0),(0:1:0),(0:0:1),(p:q:r) which line has the minimal sum of squared distances?
> > Does it pass through the centroid of the complete quadrangle?
>
> 1. Thanks Francisco, problem 2 in Coolidge article gives the answer the minimum-square-distances-line passe through the centroid of the complete quadrangle. This is also valid for all sets of n points.
> 2. I also found this phrase on 1st page in
> http://www.mamikon.com/USArticles/SumSqrsMonthly.pdf
> "Given a finite set of fixed points in 3-space, what is the locus
> of a point moving in such a way that the sum of the squares of its distances from the fixed points is constant? The answer is both elegant and surprising: the locus is a sphere whose center is at the centroid of the fixed points (if we allow the empty set and a single point as degenerate cases of a sphere)."
> This makes clear that for all sets of n points the sum of square distances is extending in circles (2-D) or spheres (3-D) with the centroid as center. So the Centroid has minimal sum of square distances to n points
> It is not the same as square distances to a line but it gives a rough idea how square distances behave with n points.
> 3. In my first calculation of the minimum-square-distances-line in the 4-point case I found also 2 perpendicular lines through the Quadrangle Centroid (similar to the Steiner Axes).
>
> Best regards,
>
> Chris van Tienhoven
>
• An obvious generalization but, what about point P(r) which has minimal sum of r-th power of distances from P(r) to sides of triangle? P(2)=Symmedian point, as
Message 6 of 19 , Jan 12, 2012
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An obvious generalization but,

what about point P(r) which has minimal sum of r-th power of distances from
P(r) to sides of triangle? P(2)=Symmedian point, as Chris commented. What
is the locus for positive r?

(note that we should take directed distances)

Regards,
Sung Hyun

[Non-text portions of this message have been removed]
• ... [snip] A symmetric equation of that line is:   x/(-2a^2+b^2+c^2+T) + y/(a^2-2b^2+c^2+T) + z/(a^2+b^2-2c^2+T) = 0 Use the symmetrization technique in
Message 7 of 19 , Jan 12, 2012
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> From: Francisco Javier <garciacapitan@...>
>
> Let's consider this question, although a long time passed since it was
> proposed.
>
> I find by calculations that the desired line goes through the centroid, and the
> line given by Martin does; however, I find another solution, with big square
> roots, if we use cartesian coordinates.
>
> With barycentric coordinates, I get a simple although  non-symmetric form of
> this line:
>
> (b^2 - c^2)x + (a^2 - b^2 - T) y + (-a^2 + c^2 + T}z = 0
>
> where T= sqrt of  (a^4 + b^4 + c^4  - a^2 b^2 - a^2 c^2 - b^2 c^2)
>
> Best regards,
>
> Francisco Javier.
>
>
>
> --- In Hyacinthos@yahoogroups.com, martin lukarevski <martinton@...>
> wrote:
>>
>> Martin wrote:
>>
>> >For a given triangle ABC find a line such that the sum
>> >AA'^2 + BB'^2 + CC'^2 is minimal where A',B',C'  are
>> >orth.projections of A,B,C on the line.

[snip]

A symmetric equation of that line is:

x/(-2a^2+b^2+c^2+T) + y/(a^2-2b^2+c^2+T) + z/(a^2+b^2-2c^2+T) = 0

Use the symmetrization technique in message 19239 on the trilinear pole of your line.
--
Barry Wolk
• Dear friends, Let L be a tangent to the circumcircle of ABC at the point P and p, q, r be the distances AA , BB , CC of the vertices A, B, C from L. If f(P) =
Message 8 of 19 , Jan 12, 2012
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Dear friends,
Let L be a tangent to the circumcircle of ABC at the point P
and p, q, r be the distances AA', BB', CC' of the vertices A, B, C from L.
If f(P) = p^2 + q^2 + r^2 then the problem to find the minimum or maximum
of f(P) seems to be a difficult problem. If ABC is isosceles
AB = AC = b and BC = a then the problem has the following solution:

If the homogeneous barycentric coordinates of P are (x : y : z) where
y = a - z and aayz + bbzx + bbxy = 0 and
A* = (aa : -2bb : -2bb) is the antipode of A then
it is obvious that
f(A) = (4bb - aa) / 2
and we can find that
f(P) = (4bb - aa)(a^4 + b^4 - 4 a^3 z + 6 a^2 z^2 - 4 a z^3 +
2 z^4))/(4 (b^2 - a z + z^2)^2)
From the derivative of f(P) in z we find that we have extrema when (
z - 2a)(2zz - 2az + aa - bb) = 0. (1)
If A'B'C', HaHbHc is the medial and orthic triangle of ABC.
From A we draw the tangent AK to the Nine point circle of ABC
and the circle (A', AK) that meets BC at N, N'
The lines AN, AN' meet again the circumcircle at Q, Q'.
If NC = z then since AK^2 = AB'.AHb = b.c.cosA / 2 = (bb + cc - aa) / 4
we have
(NA')^2 = (2bb - aa) / 4 or
4( +-(z - a/2) )^2 = 2bb - aa or
2zz - 2az + aa - bb = 0 (2) .
Since N is the trace on BC
of the point Q we conclude from (1) that the extrema of f(P) are
given at the points A, A*, Q, Q' and f(Q) = f(Q').
From f(P) we can find that f(A*) = (a^4 + 8b^4) / 2(4bb - aa)
Eliminating z from (2) and f(P) we get f(Q) = (4bb - aa)(aa + bb) / 2(3bb - aa)
We have also
f(P) = f(A) - (bb - aa)(4bb - aa)( aa + (2z - b)^2 ) / 4(zz - az + bb)^2 . . . (3)
f(P) = f(A*) - (bb - aa)(2z - a)^2 ( (4bb - aa)(2bb - aa) - bb(2z- a)^2 ) / 4(4bb - aa)(zz - az + bb)^2 . . (4)
f(P) = f(Q) + (bb - aa)(4bb - aa)(2zz - 2az + aa - bb)^2 / 4(3bb - aa)(zz - az + bb)^2 . . (5)
f(A) - f(A*) = 4bb(bb - aa) / (4bb - aa) . . . (6)

1) If a <= b or A <= pi / 3 then f(Q) <= f(P) <= f(A) or we have {min, max} at {Q, A}.
2) If bb <= aa <= 2bb or pi / 3 <= A <= pi / 2 then f(A) <= f(P) <= f(Q) and {min, max} at {A, Q}
3) If 2bb <= aa < 4bb or pi / 2 <= A < pi then there is no Q and from
f(A) <= f(P) <= f(A*) we have that {min, max} at {A, A*}.

Best regards
• Sorry for a typo. I my previou message I wrote that the distances p, q, r are AA , BB , CC correct to AA , BB , CC or delete it because it is not used in the
Message 9 of 19 , Jan 12, 2012
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Sorry for a typo.
I my previou message
I wrote that the distances
p, q, r are AA', BB', CC'
correct to AA", BB", CC"
or delete it because it is
not used in the sequel
and A'B'C' is the medial triangle.
ND

> Dear friends,
> Let L be a tangent to the circumcircle of ABC at the point
> P
> and p, q, r be the distances AA', BB', CC' of the vertices
> A, B, C from L.
> If f(P) = p^2 + q^2 + r^2 then the problem  to find
> the minimum or maximum
> of f(P) seems to be a difficult problem. If ABC is
> isosceles
> AB = AC = b and BC = a then the problem has the following
> solution:
>
> If the homogeneous barycentric coordinates of P are (x : y
> : z) where
> y = a - z  and  aayz + bbzx + bbxy = 0  and
>
> A* = (aa : -2bb : -2bb)  is the antipode of A
> then
> it is obvious that
> f(A) = (4bb - aa) / 2
> and we can find that
> f(P) = (4bb - aa)(a^4 + b^4 - 4 a^3 z + 6 a^2 z^2 - 4 a z^3
> +
>     2 z^4))/(4 (b^2 - a z + z^2)^2)
> From the derivative of f(P) in z we find that we have
> extrema when (
> z - 2a)(2zz - 2az + aa - bb) = 0. (1)
> If A'B'C', HaHbHc  is the medial and orthic triangle
> of ABC.
> From A we draw the tangent AK to the Nine point circle of
> ABC
> and the circle (A', AK) that meets BC at N, N'
> The lines AN, AN' meet again the circumcircle at Q, Q'.
> If  NC = z  then since AK^2 = AB'.AHb = b.c.cosA
> / 2 = (bb + cc - aa) / 4
> we  have
> (NA')^2 = (2bb  - aa) / 4    or
> 4( +-(z - a/2) )^2 = 2bb - aa   or
> 2zz - 2az + aa - bb = 0  (2) .
> Since N is the trace on BC
> of the point Q we conclude from (1) that the extrema of
> f(P) are
> given at the points A, A*, Q, Q' and f(Q) = f(Q').
> From f(P) we can find that f(A*) = (a^4 + 8b^4) / 2(4bb -
> aa)
> Eliminating z from (2) and f(P)  we get f(Q) = (4bb -
> aa)(aa + bb) / 2(3bb - aa)
> We have  also
> f(P) = f(A) - (bb - aa)(4bb - aa)( aa + (2z - b)^2 ) / 4(zz
> - az + bb)^2 . . .  (3)
> f(P) = f(A*) - (bb - aa)(2z - a)^2 ( (4bb - aa)(2bb -
> aa)  - bb(2z- a)^2 ) / 4(4bb - aa)(zz - az + bb)^2 . .
> (4)
> f(P) = f(Q) + (bb - aa)(4bb - aa)(2zz - 2az + aa - bb)^2 /
> 4(3bb - aa)(zz - az + bb)^2  . . (5)
> f(A) - f(A*) = 4bb(bb - aa) / (4bb - aa) . . . (6)
>
> 1) If a <=  b  or  A <= pi / 3
> then f(Q) <= f(P) <= f(A)  or we have {min, max}
> at {Q, A}.
> 2) If bb <= aa <= 2bb or pi / 3 <= A <= pi / 2
> then  f(A) <= f(P) <= f(Q)   and
> {min, max} at {A, Q}
> 3) If 2bb <= aa < 4bb or pi / 2 <= A < pi
> then there is no Q and from
>    f(A) <= f(P) <= f(A*)  we have
> that {min, max} at {A, A*}.
>
> Best regards
>
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>
• Nikolaos, Thank you for using and enyoing Baricentricas.nb, and for your solution that I ll read later. note also that there is a function
Message 10 of 19 , Jan 12, 2012
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Nikolaos,

Thank you for using and enyoing Baricentricas.nb, and for your solution

the distance from a point to a line.

> Sorry for a typo.
> I my previou message
> I wrote that the distances
> p, q, r are AA', BB', CC'
> correct to AA", BB", CC"
> or delete it because it is
> not used in the sequel
> and A'B'C' is the medial triangle.
> ND
>
> > Dear friends,
> > Let L be a tangent to the circumcircle of ABC at the point
> > P
> > and p, q, r be the distances AA', BB', CC' of the vertices
> > A, B, C from L.
> > If f(P) = p^2 + q^2 + r^2 then the problem to find
> > the minimum or maximum
> > of f(P) seems to be a difficult problem. If ABC is
> > isosceles
> > AB = AC = b and BC = a then the problem has the following
> > solution:
> >
> > If the homogeneous barycentric coordinates of P are (x : y
> > : z) where
> > y = a - z and aayz + bbzx + bbxy = 0 and
> >
> > A* = (aa : -2bb : -2bb) is the antipode of A
> > then
> > it is obvious that
> > f(A) = (4bb - aa) / 2
> > and we can find that
> > f(P) = (4bb - aa)(a^4 + b^4 - 4 a^3 z + 6 a^2 z^2 - 4 a z^3
> > +
> > 2 z^4))/(4 (b^2 - a z + z^2)^2)
> > From the derivative of f(P) in z we find that we have
> > extrema when (
> > z - 2a)(2zz - 2az + aa - bb) = 0. (1)
> > If A'B'C', HaHbHc is the medial and orthic triangle
> > of ABC.
> > From A we draw the tangent AK to the Nine point circle of
> > ABC
> > and the circle (A', AK) that meets BC at N, N'
> > The lines AN, AN' meet again the circumcircle at Q, Q'.
> > If NC = z then since AK^2 = AB'.AHb = b.c.cosA
> > / 2 = (bb + cc - aa) / 4
> > we have
> > (NA')^2 = (2bb - aa) / 4 or
> > 4( +-(z - a/2) )^2 = 2bb - aa or
> > 2zz - 2az + aa - bb = 0 (2) .
> > Since N is the trace on BC
> > of the point Q we conclude from (1) that the extrema of
> > f(P) are
> > given at the points A, A*, Q, Q' and f(Q) = f(Q').
> > From f(P) we can find that f(A*) = (a^4 + 8b^4) / 2(4bb -
> > aa)
> > Eliminating z from (2) and f(P) we get f(Q) = (4bb -
> > aa)(aa + bb) / 2(3bb - aa)
> > We have also
> > f(P) = f(A) - (bb - aa)(4bb - aa)( aa + (2z - b)^2 ) / 4(zz
> > - az + bb)^2 . . . (3)
> > f(P) = f(A*) - (bb - aa)(2z - a)^2 ( (4bb - aa)(2bb -
> > aa) - bb(2z- a)^2 ) / 4(4bb - aa)(zz - az + bb)^2 . .
> > (4)
> > f(P) = f(Q) + (bb - aa)(4bb - aa)(2zz - 2az + aa - bb)^2 /
> > 4(3bb - aa)(zz - az + bb)^2 . . (5)
> > f(A) - f(A*) = 4bb(bb - aa) / (4bb - aa) . . . (6)
> >
> > 1) If a <= b or A <= pi / 3
> > then f(Q) <= f(P) <= f(A) or we have {min, max}
> > at {Q, A}.
> > 2) If bb <= aa <= 2bb or pi / 3 <= A <= pi / 2
> > then f(A) <= f(P) <= f(Q) and
> > {min, max} at {A, Q}
> > 3) If 2bb <= aa < 4bb or pi / 2 <= A < pi
> > then there is no Q and from
> > f(A) <= f(P) <= f(A*) we have
> > that {min, max} at {A, A*}.
> >
> > Best regards
> >
> >
> >
> >
> > ------------------------------------
> >
> >
> >
> > Hyacinthos-fullfeatured@yahoogroups.com
> >
> >
> >
>
>
> ------------------------------------
>
>
>
>
>

--
---
Francisco Javier García Capitán
http://garciacapitan.99on.com

[Non-text portions of this message have been removed]
• Dear friends, So X(6) in a triangle has minimal sum of squared distances to the sides of a reference triangle. I wondered which point Q in a complete
Message 11 of 19 , Jan 16, 2012
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Dear friends,

So X(6) in a triangle has minimal sum of squared distances to the sides of a reference triangle.
I wondered which point Q in a complete quadrilateral has minimal sum of squared distances to the sides of the complete quadrilateral.
To calculate this I had to modify the construction method of J.L. Coolidge:
Let O (origin), A and B be random non-collinear points.
Go = Quadrangle-Centroid of projection points of O on the basic 4 lines of the Reference Quadrilateral.
Ga = Quadrangle-Centroid of projection points of O on the 4 lines through point A parallel to the 4 basic lines of the Reference Quadrilateral.
Gb = Quadrangle-Centroid of projection points of O on the 4 lines through point B parallel to the 4 basic lines of the Reference Quadrilateral.
Let Sa = Ga.Go ^ O.B and Sb = Gb.Go ^ O.A.
Construct A1 on line O.A such that Sa.Ga : Ga.Go = O.A : A.A1.
Construct B1 on line O.B such that Sb.Gb : Gb.Go = O.B : B.B1.

Construct Q such that O.A1.P.B1 is a parallelogram where O and P are opposite vertices. Now Q is the point of Least Squares.

Amazing in this method is that starting points O, A, B can be chosen randomly. This makes it possible to calculate the point Q by taking simple values for these points. This has no consequence for the outcome.
Calculation when the 4 basic lines of the complete quadrilateral are:
L1=(1:0:0), L2=(0:1:0), L3=(0:0:1), L4=(l:m:n),
gives 1st coordinate of Q:
a^2 (a^2(l-m)(l-n) + b^2(2m-n)(m-l) + c^2(2n-m)(n-l)).
The 2nd and 3rd coordinates are determined cyclically.
Just a bit more complicated than the 1st coordinate of X(6): a^2.

Best regards,

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" <van10hoven@...> wrote:
>
> Dear Friends,
>
> The line with minimal sum of squared distances to Triangle vertices is related to the point with minimal sum of squared distances to the sides of a triangle (duality principle).
> Following the instructions from the paper of J.L. Coolidge I figured out which point in a triangle has minimal sum of squared distances to the sides of a reference triangle.
> It is not the X2 (centroid), it is not X13 (Fermat Point).
> It is X6 (Symmedian Point). I wouldn't have expected this.
>
> Best regards,
>
> Chris van Tienhoven
• Dear Friends, In my last message about the construction of the point with minimal sum of squared distances to the 4 lines of a complete quadrilateral I made a
Message 12 of 19 , Jan 25, 2012
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Dear Friends,

In my last message about the construction of the point with minimal sum of squared distances to the 4 lines of a complete quadrilateral I made a mistake.
I mentioned:
Let Sa = Ga.Go ^ O.B and Sb = Gb.Go ^ O.A.
This should be:
Let Sa = Ga.Go ^ O.Gb and Sb = Gb.Go ^ O.Ga.
I am sorry for the inconvenience.
best regards,

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" <van10hoven@...> wrote:
>
> Dear friends,
>
> So X(6) in a triangle has minimal sum of squared distances to the sides of a reference triangle.
> I wondered which point Q in a complete quadrilateral has minimal sum of squared distances to the sides of the complete quadrilateral.
> To calculate this I had to modify the construction method of J.L. Coolidge:
> Let O (origin), A and B be random non-collinear points.
> Go = Quadrangle-Centroid of projection points of O on the basic 4 lines of the Reference Quadrilateral.
> Ga = Quadrangle-Centroid of projection points of O on the 4 lines through point A parallel to the 4 basic lines of the Reference Quadrilateral.
> Gb = Quadrangle-Centroid of projection points of O on the 4 lines through point B parallel to the 4 basic lines of the Reference Quadrilateral.
> Let Sa = Ga.Go ^ O.B and Sb = Gb.Go ^ O.A.
> Construct A1 on line O.A such that Sa.Ga : Ga.Go = O.A : A.A1.
> Construct B1 on line O.B such that Sb.Gb : Gb.Go = O.B : B.B1.
>
> Construct Q such that O.A1.P.B1 is a parallelogram where O and P are opposite vertices. Now Q is the point of Least Squares.
>
> Amazing in this method is that starting points O, A, B can be chosen randomly. This makes it possible to calculate the point Q by taking simple values for these points. This has no consequence for the outcome.
> Calculation when the 4 basic lines of the complete quadrilateral are:
> L1=(1:0:0), L2=(0:1:0), L3=(0:0:1), L4=(l:m:n),
> gives 1st coordinate of Q:
> a^2 (a^2(l-m)(l-n) + b^2(2m-n)(m-l) + c^2(2n-m)(n-l)).
> The 2nd and 3rd coordinates are determined cyclically.
> Just a bit more complicated than the 1st coordinate of X(6): a^2.
>
> Best regards,
>
> Chris van Tienhoven
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