A square has an area of 10 cm^2. Two straight lines are drawn across the square dividing it into four areas of 1 cm^2, 2 cm^2, 3 cm^2, and 4 cm^2.
Where can those lines be drawn?
Since 1+4 = 2+3, the first line can divide the square into two equal halves. Experimentation with Geometer's Sketchpad shows that a diagonal of the square, or a line parallel to two of the edges will allow adjustment of a second line to divide one side into a 1:4 ratio and the other into a 2:3 ratio. I'm sure there are (infinitely) many others.
The easier case to get more exact numbers would be the latter case as the four resulting pieces are trapezoids.
The playing around with GSP shows that if the halfway dividing line is horizontal, then if the transverse line is 1/sqrt(10) to the right of the left side on the top edge of the square and crosses the bottom edge at its center, the areas will be:
Top left: 5 * (1+3)/2 / 10 = 1
Top right: 5 * (9+7)/2 / 10 = 4
Bottom left: 5 * (3+5)/2 / 10 = 2
Bottom right: 5 * (7+5)/2 / 10 = 3
Playing around with the diagonal cut in GSP shows a more complicated situation: If the diagonal is from top left to bottom right, then a line from top to bottom that cuts the top edge of the square about 48.3% of the way from left to right, and cuts the bottom edge about 31.8% of the way from left to right, would result in a triangle at upper-left of area 1, a triangle at lower right with area 2, a quadrilateral of area 3 at the bottom-left and a quadrilateral of area 4 at the upper right.
The percentages of course are of sqrt(10).
It doesn't seem possible to have the two transverse lines be perpendicular to each other.
Posted by Charlie
on 2016-08-12 10:38:54