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Line Dividing a Plane

Date: 05/11/2001 at 01:23:53
From: Jessica Holmes
Subject: A line dividing a plane

Dear Dr. Math,

I am given a square (graphed on the Cartesian coordinate system) and a 
point in the square, and I must draw a line through the point that 
will divide the square into two regions: one being the smallest area 
possible, and the other being the largest possible. I don't know how 
the line should be drawn. What should be the slope of the line? Is 
there a theorem for it? How would you prove it?

I tried guess and check with specific cases. I used numbers for the 
coordinates of the given point, and I tried a variety of slopes for 
each case. The best results always seem to be the value of x times the 
value of y times 2. I think it has something to do with the rectangle 
that is formed by drawing the perpendicular lines from the given point 
to the x and y axes. I drew a diagonal from the given point to the 
origin. I arranged the two triangles on two sides of the square to 
form the smallest-area region. I believe this works but I do have 
doubts because I don't have any justification.

Date: 05/11/2001 at 13:28:24
From: Doctor Rob
Subject: Re: A line dividing a plane

Thanks for writing to Ask Dr. Math, Jessica.

I will assume henceforth that the square has sides parallel to the 
coordinate axes, and has corners (x1,y1), (x1,y2), (x2,y1), and 
(x2,y2), where x1 < x2, y1 < y2. Let the point in question be (x0,y0).

There are four cases, depending on which corner of the square is
closest to (x0,y0). We will assume that it is (x1,y1), the lower left
corner. Then the line through (x0,y0) with slope m has equation

   y = m*x + y0 - m*x0.

We will assume that the line cutting the smallest area has m < 0, and
that it intersects the lower and left sides of the square (so that the
cut-off piece is a right triangle and not a trapezoid).  Then the area
of the intersected right triangle is half the product of its legs, or

   A = [-(x0-x1)*m+(y0-y1)]*[-(y0-y1)/m+(x0-x1)]/2.

For the sake of simplicity, let's put X = x0 - x1 and Y = y0 - y1.  

   A = (1/2)*(-X*m+Y)*(-Y/m+X).

The trick is to rewrite this in such a way as to make the minimum
value obvious:

   A = (1/2)*(2*X*Y - X^2*m - Y^2/m),
     = (1/2)*(4*X*Y + X^2*[-m] - 2*X*Y + Y^2/[-m]),
     = 2*X*Y + (1/2)*(X*sqrt[-m]-Y/sqrt[-m])^2.

This has the form c + (1/2)*u^2, for some constant c. This is minimum
when u = 0, because squares are always >= 0. Set the square equal to
zero and solve for m to find the best value of m. Then the minimum
area is the constant c, that is,

   A_min = 2*X*Y = 2*(x0-x1)*(y0-y1).

I leave it to you to determine what happens if it is a corner other
than the lower left one that is closest to (x0,y0). There are three
cases to consider, all of which are similar to the above analysis.

If there are two corners tied for closest to (x0,y0), then it lies on
one of the lines, either vertical or horizontal, which pass through 
the center of the square, that is, x = (x2+x1)/2 or y = (y2+y1)/2. In 
this case, the best line is either y = y0 or x = x0, respectively, and 
the minimum area is either

   A_min = (x2-x1)*min(y2-y0,y0-y1)/2
   A_min = min(x2-x0,x0-x1)*(y2-y1)/2,

respectively, and the smallest piece is a rectangle.

If all four corners are tied for closest, then (x0,y0) is the center 
of the square, and any line through it will divide the square into two
congruent pieces.  That means no line is better than any other, and 
the minimum area is (x2-x1)*(y2-y1)/2.

- Doctor Rob, The Math Forum   
Associated Topics:
High School Coordinate Plane Geometry
High School Equations, Graphs, Translations
High School Geometry
High School Triangles and Other Polygons

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