Triangle-Minimum Distance
Let's consider an acute angle triangle $ (ABC )$.
Let's determine a point $P$ (in the plane of the triangle) such that the sum of the distances from $P $ to the vertexes $(A, B, C )$ is minimum.
Is this point unique ?
Let's determine a point $P$ (in the plane of the triangle) such that the sum of the distances from $P $ to the vertexes $(A, B, C )$ is minimum.
Is this point unique ?
Risposte
Denote by f the rotation through 60° counterclockwise about A. For any point
M in the plane let M' = f(M). Then AMM' is an equilateral triangle.
Consider an arbitrary point X in ABC. Then AX = XX', while X'= f(X)
and f(C) = C' imply CX = C'X'. Consequently, AX+BX+CX=XX'+BX+C'X' >BC'.
Denote by Xo the intersection point of BC' with the
circumcircle of ACC' and X'o=f(Xo)
Moreover, we have
AXo+BXo+CXo=XoX'o+BXo+C'X'o=BC'
Let’s name the 3 vertexes of the acute angle triangle as follows :
$vecp_1 =(x_1,y_1) ; vecp_2=(x_2,y_2) ; vecp_3=(x_3,y_3)$
and $vecp=(x,y)$ being the point to determine , which minimizes the sum of the distances from the 3 vertexes.
The objective function $(f : RR^2 -> RR ) $ to be minimized is given by $f(x,y)=sum_(i=1)^3 d_i(x,y)$ where $d_i(x,y)= |vecp-vecp_i | = sqrt((x_i-x)^2+(y_i-y)^2)$ for $i=1,2,3 $.
If the minimum point exists , it will lie inside a closed circle C of ray large enough, while certainly the function has no maximum.
We have then to look for the point in C , closed and bounded set of $RR^2$.
Since $ f $ is a continuous function, Weierstrass theorem assures the existence of a point $vecp $ for which $f $ has the global minimum.
Therefore the problem has solution .
The function $ f $ is strictly convex, therefore the minimum point exists and is unique.
Proof of function $f $ being strictly convex in the plane $xy$.
***************************************************************************
Each addendum is convex, not strictly, since, for each $t in (0,1)$and each pair of points $vecp, vecq$ we have :
$|tvecp+(1-t)vecq-vecp_i|=|t(vecp-vec p_i)+(1-t)(vecq-vecp_i)| <= |vecp-vecp_i|+(1-t)|vec q-vecp_i|.
The sign $= $ is applicable only in the case where vectors $vecp-vecp_i$ and $vecq-vec vecp_i $ are linearly dependent.
In this case we must have, for every $vec p,vecq,vecp ne vec q $ :
$f(tvecp+(1-t)vecq)=sum_(n=1)^3|tvecp+(1-t)vecq-vecp_i|
*****************************************************************************
Therefore $f $ is strictly convex and in this case a theorem assures that the minimum point is unique.
Let’s look for the minimum point : $f $ is differentiable in all points of the plane $xy $ except in the vertexes of the triangle.
We verify that none of these points can be the minimum point.
Let’s in fact consider the vertex common to the two shortest sides of the triangle.
Let $vecp_1$ be this vertex and $vecq$ the foot of the perpendicular line from vertex $vecp_2 $ to the opposite side .
It is easy to verify that
$f(vecq)=|vecq-vecp_1|+|vecq-vecp_2|+|vecq-vecp_3| =|vecq-vecp_2| + | vecp_1-vecp_3| <= | vecp_1-vecp_2|+|vecp_1-vecp_3|=f(vecp_1) $ and consequently $vecp_1 $ cannot be the minimum point.
Let’s now consider the other points of the plane ( not the vertexes).
Let’s look for the critical points , equalling $grad f $ to $ 0 $.
We get the following system :
$(delf)/(delx) =(x-x_1)/(d_1(x,y) )+(x-x_2)/(d_2(x,y))+(x-x_3)/(d_3(x,y)) =0 $
$(delf)/(dely) =(y-y_1)/(d_1(x,y) )+(y-y_2)/(d_2(x,y))+(y-y_3)/(d_3(x,y)) =0 $
For solving the system we introduce 3 unit vectors :
$w_1=((x-x_1)/(d_1(x,y)), (y-y_1)/(d_1(x,y))) $ ; $ w_2==((x-x_2)/(d_2(x,y)), (y-y_2)/(d_2(x,y))) $ ;
$w_3= ((x-x_3)/(d_3(x,y)), (y-y_3)/(d_3(x,y))) $ .
We note that the system is equivalent to the following vector equation :
$w_1+w_2+w_3 =0 $
Three unit vectors not null in the plane can have as sum the null vector only if the corresponding points of the plane are situated at the vertexes of an equilateral triangle , that is the angle that each one forms with the other two is equal to 120° .
Only one point satisfies this condition and it is internal to the triangle.
The standard approach, calculating hessian matrix of $ f $ and evaluating its sign is much more complicated.
$vecp_1 =(x_1,y_1) ; vecp_2=(x_2,y_2) ; vecp_3=(x_3,y_3)$
and $vecp=(x,y)$ being the point to determine , which minimizes the sum of the distances from the 3 vertexes.
The objective function $(f : RR^2 -> RR ) $ to be minimized is given by $f(x,y)=sum_(i=1)^3 d_i(x,y)$ where $d_i(x,y)= |vecp-vecp_i | = sqrt((x_i-x)^2+(y_i-y)^2)$ for $i=1,2,3 $.
If the minimum point exists , it will lie inside a closed circle C of ray large enough, while certainly the function has no maximum.
We have then to look for the point in C , closed and bounded set of $RR^2$.
Since $ f $ is a continuous function, Weierstrass theorem assures the existence of a point $vecp $ for which $f $ has the global minimum.
Therefore the problem has solution .
The function $ f $ is strictly convex, therefore the minimum point exists and is unique.
Proof of function $f $ being strictly convex in the plane $xy$.
***************************************************************************
Each addendum is convex, not strictly, since, for each $t in (0,1)$and each pair of points $vecp, vecq$ we have :
$|tvecp+(1-t)vecq-vecp_i|=|t(vecp-vec p_i)+(1-t)(vecq-vecp_i)| <= |vecp-vecp_i|+(1-t)|vec q-vecp_i|.
The sign $= $ is applicable only in the case where vectors $vecp-vecp_i$ and $vecq-vec vecp_i $ are linearly dependent.
In this case we must have, for every $vec p,vecq,vecp ne vec q $ :
$f(tvecp+(1-t)vecq)=sum_(n=1)^3|tvecp+(1-t)vecq-vecp_i|
*****************************************************************************
Therefore $f $ is strictly convex and in this case a theorem assures that the minimum point is unique.
Let’s look for the minimum point : $f $ is differentiable in all points of the plane $xy $ except in the vertexes of the triangle.
We verify that none of these points can be the minimum point.
Let’s in fact consider the vertex common to the two shortest sides of the triangle.
Let $vecp_1$ be this vertex and $vecq$ the foot of the perpendicular line from vertex $vecp_2 $ to the opposite side .
It is easy to verify that
$f(vecq)=|vecq-vecp_1|+|vecq-vecp_2|+|vecq-vecp_3| =|vecq-vecp_2| + | vecp_1-vecp_3| <= | vecp_1-vecp_2|+|vecp_1-vecp_3|=f(vecp_1) $ and consequently $vecp_1 $ cannot be the minimum point.
Let’s now consider the other points of the plane ( not the vertexes).
Let’s look for the critical points , equalling $grad f $ to $ 0 $.
We get the following system :
$(delf)/(delx) =(x-x_1)/(d_1(x,y) )+(x-x_2)/(d_2(x,y))+(x-x_3)/(d_3(x,y)) =0 $
$(delf)/(dely) =(y-y_1)/(d_1(x,y) )+(y-y_2)/(d_2(x,y))+(y-y_3)/(d_3(x,y)) =0 $
For solving the system we introduce 3 unit vectors :
$w_1=((x-x_1)/(d_1(x,y)), (y-y_1)/(d_1(x,y))) $ ; $ w_2==((x-x_2)/(d_2(x,y)), (y-y_2)/(d_2(x,y))) $ ;
$w_3= ((x-x_3)/(d_3(x,y)), (y-y_3)/(d_3(x,y))) $ .
We note that the system is equivalent to the following vector equation :
$w_1+w_2+w_3 =0 $
Three unit vectors not null in the plane can have as sum the null vector only if the corresponding points of the plane are situated at the vertexes of an equilateral triangle , that is the angle that each one forms with the other two is equal to 120° .
Only one point satisfies this condition and it is internal to the triangle.
The standard approach, calculating hessian matrix of $ f $ and evaluating its sign is much more complicated.
Yes, later I will post an analytical solution.
As reference I indicate :
http://en.wikipedia.org/wiki/Fermat_point
As reference I indicate :
http://en.wikipedia.org/wiki/Fermat_point
@Camillo: Will you ever post a solution?
I didn't succeed in finding an analytical solution.
Recently I've read somewhere this problem is kinda classic, for it dates back to J. Steiner (about 170 years ago); it seems that Steiner himself solved it (using some geometric tools as he was used to do for almost any type of problem!
) and, if my memory doesn't fail, the solution is the point [tex]$P$[/tex] in the triangle s.t. it views each side under an angle of [tex]$120^\circ$[/tex].
Is it true?
I didn't succeed in finding an analytical solution.
Recently I've read somewhere this problem is kinda classic, for it dates back to J. Steiner (about 170 years ago); it seems that Steiner himself solved it (using some geometric tools as he was used to do for almost any type of problem!
) and, if my memory doesn't fail, the solution is the point [tex]$P$[/tex] in the triangle s.t. it views each side under an angle of [tex]$120^\circ$[/tex].Is it true?
Ok ziomauri! That's a nice proof.
I'd like to remark that, with the aid of a (not so) much complicated reasoning, the following more precise statement can be proved:
For each point [tex]$X$[/tex] which doesn't lie in the triangle [tex]$ABC$[/tex] nor on its sides, there exists at least one point [tex]$X^\prime$[/tex] on the closest [tex]$ABC$[/tex]'s side to [tex]$X$[/tex] s.t. [tex]$|AX^\prime|+|BX^\prime|+|CX^\prime| < |AX|+|BX|+|CX|$[/tex].
The trick consists in using the elementary fact that in a triangle the longest side is always opposite to the widest inner angle.
I'd like to remark that, with the aid of a (not so) much complicated reasoning, the following more precise statement can be proved:
For each point [tex]$X$[/tex] which doesn't lie in the triangle [tex]$ABC$[/tex] nor on its sides, there exists at least one point [tex]$X^\prime$[/tex] on the closest [tex]$ABC$[/tex]'s side to [tex]$X$[/tex] s.t. [tex]$|AX^\prime|+|BX^\prime|+|CX^\prime| < |AX|+|BX|+|CX|$[/tex].
The trick consists in using the elementary fact that in a triangle the longest side is always opposite to the widest inner angle.
It is easy to see that if X is outside ABC, then there exists a point X'
such that AX'+BX'+CX' < AX+BX+CX. Indeed, suppose that X is exterior to the triangle. Then one
of the lines AB, BC, CA, say AB, has the property that ABC and the point X
lie in different half-planes determined by this line.
Consider the reflection X' of X in AB. We have AX' = AX, BX' = BX. Also,
the line segment CX intersects the line AB at some point Y , and XY = X'Y. Now
the triangle inequality gives CX'< CY + X'Y = CY + XY = CX,implying
AX'+BX'+CX' < AX+BX+CX
By means of Elementary Geometry, prove that the point $P$ cannot lie outside the triangle.
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