Geometry Forum - Problem of the Week Solutions - Dividing the Land, May 23-27, 1994 Susan Quan ____________________ Jamal's claim is true. By using Sketchpad I found that the products of the areas are equal, but this is not a fair way to divide the land since the amount of land is the sum not the product of the pieces. Again using Sketchpad, I found that if the midpoints of the opposite sides of a quadrilateral are drawn, the sum of the areas of regions in upper right plus lower left = sum of areas of regions in lower right plus upper left. Alan Schultz ____________________ Jamal's claim is true about when the products of the areas is equal. But it is not a fair deal, because they are not equal. Instead of connecting opposite vertices with intersecting diagonals, connect midpoints of AB and BC with those of CD and AD. [Extra] This comes out to be fair. Gino Perrotte ____________________ Again, I began the problem by graphing it. Each unit on the graph paper is worth 1. This is an easy way to find the length of the segments making up the quadrilateral. Using this method, AB = 2.75, BC = 5, CD = 4, AD = 5.75. I divided up the quadrilateral by the given directions. Four triangles were formed. I wanted to find the area of each of these triangles. To do this you must insert the altitudes of each triangle, and then measure them on the graph paper. The point of intersection of the diagonals AC and DB is called point E. The altitude of CD is found to be 3.25, the altitude of AD is 1.5, the altitude of AB is 1.5, and the altitude of BC is 1.25. Then I found the area of each of the triangles by the formula area = .5base*height. Doing the math the answers come out to be area of triangle AEB is 2.1 units squared. The area of triangle DEC is 6.5 units squared. The area of triangle AED is 4.3 units squared. The area of triangle BEC is 3.1 units squared. Then add together the area of triangles AEB and DEC and you get a total of 8.6 units squared. Then add together the area of triangles AED and BEC and you get a total of 7.4 units squared. Therefore the areas are not equal and Jacob was cheated. Also, I found a way to divide a quadrilateral up easily by using three segments. First draw a diagonal. Then from the two vertices left over, draw a segment from each of the vertices to the midpoint of the diagonal. There are four triangles formed. Two have a measure equal to each other, and the other two have an area equal to each other. Add one of each pair together and you wind up with a quadrilateril with two sections congruent. Rob Gallagher and Ryan Ferchak ____________________ We used the Geometer's Sketchpad and constructed two different models, calculated the four areas, then added pairs and multiplied pairs. Here are our conclusions. Jamal's claim is true about the products being equal. This way is not a fair way to divide the land. Jacob was right; the sums were not equal and that's all that matters when dividing the land equally. Susie Taylor and Valerie Stalter ____________________ Jamal's claim that the products in this quadrilateral are equal, and that this is a fair way to divide the land, would only work if the quadrilateral were a square or rectangle. In an irregular quadrilateral the triangles formed by the diagonals are not necessarily congruent. The triangles formed could have equal products without having equal sums; for example if triangle BEA had an area of 3, and CDE had an area of 4, then the sum of these would be 7 and the products would be 12. In triangle BEC the area is 1, in AED the area is 12; then the sum is 13, and the product is 12. This shows that the products can be equal without the sums being equal, and by the area addition postulate the total area is found by the sum, not the product of the areas. Extra: Connect the midpoints of one pair of opposite sides, then draw the perpendicular bisector to the line. The sums of the non- consecutive quadrilaterals sharing vertical angles would then be equal. Robert Dawson ____________________ Here's a solution to the 'Jamal & Jacob' problem which was posted. It may not be the simplest, but it illustrates a technique which I have often found useful. _Affine_transformations_preserve_the_ratios_of_areas_. Thus, without loss of generality, we may assume the diagonals of the piece of land to be orthogonal. Then the areas of the four triangular plots are just (AE)(BE)/2, (BE)(CE)/2, (CE)(DE)/2, and (DE)(AE)/2; and the product of the areas of either contractor's two lots is (AE)(BE)(CE)(DE)/4. Of course, any quadrilateral may have its area divided into two equal parts by *one* line segment ;-> But this generally requires some calculation to place, although there are infinitely many ways to do it. We may suppose that J&J went to a school in which arithmetic was not encouraged, and that their calculator batteries are dead; and that they want the geometry to automagically do the calculation for them. Very well; let them join opposite pairs of _midpoints_, and again take opposite lots. This will provide a fair [sums of areas are equal] division. To see this, let W be the midpoint of AB, X of BC, Y of CD, and Z of DA. Let F be the intersection of the two segments WY and XZ. It is clear that triangles AWF and BWF have equal bases AW and BW, and the same height, so their areas are the same. This is also true of (BXF,CXF),(CYF,DYF), and (DZF,AZF). Now, Jacob gets (say) AWFZ and DYFX, which may be subdivided into the triangles AWF,AZF, CXF, and CYF - one triangle from each pair. Jamal gets BXFW and DYFZ, which between them comprise the other triangle from each pair. As each pair of triangles are equal, J&J should be satisfied. Question: Can this be simplified at all by the use of _projective_ transformations taking the plot ABCD onto a square? (Probably not!) Previous page || Next problem || Previous problem || Table of Contents || Forum Home Page