Line Dividing a PlaneDate: 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. Then 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 or 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 http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/