Lesson The point which minimizes the sum of distances to vertices of a given quadrilateral

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<H2>The point which minimizes the sum of distances to vertices of a given quadrilateral</H2>

<H3>Problem 1</H3>Find the smallest possible value of an expression &nbsp;&nbsp;{{{sqrt(x^2+y^2)}}} + {{{sqrt((x-1)^2+y^2)}}} + {{{sqrt(x^2+(y-1)^2)}}} + {{{sqrt((x-3)^2+(y-4)^2)}}}.

<B>Solution</B>

The key to the solution of the problem is to notice that the given expression is the sum of distances from the point &nbsp;(x,y)
in a coordinate plane to the points &nbsp;(0,0), &nbsp;(0,1), &nbsp;(1,0) &nbsp;and &nbsp;(3,4).

<U>Theorem</U>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;For a convex quadrilateral in a plane, the point in the plane which minimizes the sum of the distances from the point to vertices 
    &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;of the quadrilateral is the intersection of its diagonals.

<pre>
          Similar statement for a triangle leads to Fermat's point of a triangle and is considered as a difficult geometry
          conception and statement, which goes far beyond and above the elementary geometry level.

See this Wikipedia article  https://en.wikipedia.org/wiki/Fermat_point .

But for a quadrilateral it is ELEMENTARY statement accesible and approachable for starters.
</pre>

<U>Proof</U>

<pre> 
Let ABCD be the given quadrilatersl in a plane with the verices A, B, C and D  (in this order).

Let "O" be the intersection point of its diagonals AC and BD.

And let X be any other point in the plane.

The sum of distances from X to vertices is

d(X) = |AX| + |BX| + |CX| + |DX|.

The sum of distances from O to vertices is

d(O) = |AO| + |BO| + |CO| + |DO|.

By applying the "triangle inequality", you have

d(O) = (|AO| + |CO|) + (|BO| + |DO|) = |AC| + |BD| < (|AX| + |CX|) + (|BX| + |DX|) = d(X),

and the statement is PROVED.
</pre>

Therefore, &nbsp;the solution to the problem is &nbsp;<U>THIS</U>:

<pre>
    The point in the plane which gives the minimum to your expression is the intersection point of the segment
    connecting the points A=(0,0) and C=(3,4) with the segment connecting the points B=(0,1) and D=(1,0).

The straight line connecting the points A and C is

y = {{{(4/3)*x}}}.     (1)

The straight line connecting the points B and D is

y - 1 = -x.      (2)

Their intersection is the point

{{{(4/3)*x}}} = 1 - x  ====>  4x = 3 - 3x  ====>  7x = 3  ====>  x = {{{3/7}}};  y = {{{(4/3)*(3/7)}}} = {{{4/7}}}.

To find the minimum of the given expression, you need to find the lengths of the diagonals  |AC| and |BD|  and add them.

|AC| = {{{sqrt((3-0)^2 + (4-0)^2)}}} = {{{sqrt(3^2 + 4^2)}}} = {{{sqrt(25)}}} = 5;

|BD| = {{{sqrt(2)}}}.

So, the minimum of the given expression is  {{{5 + sqrt(2)}}}.

<U>Answer</U>.  The point which provides the minimum to the given expression is (x,y) = ({{{3/7)}}},{{{4/7}}}).

The value of the minimum is {{{5 + sqrt(2)}}}.
</pre>

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