Geometry Forum - Problem of the Week Solutions - Congruent Squares, April 25-29, 1994 Mark Plesko ____________________ [This is my ascii interpretation of his drawing sent via NeXT mail. -Annie] _____________ | | | | As the lower square rotates, the value of | | changes. The area of the overlapping region is | ^ | | ' | \ | A = 3x + 2((1/2)(3)(3-x)) | ' |3-x | = 3x + 9 - 3x |'_____| \ | = 9 '|______|x___\_| ' 3 3-x \ No matter what the angle is, the area is \ \ 9 square cm. \ ' \ ' \ ' \ ' \ ' \ ' Billy Glisson, Billy Becht, Tim Pappas, and Fernando Davila ____________________ First we worked with 2 squares(6cm x 6cm). We found the center by drawing the two diagonals and were they intercected was the center of the square. Then we took the vertex of the other square and put it in the center of the square so they overlaped. We found that they overlaped by a 3 x 3 cm square which equaled a area of 9 cm squared. We turned the square to make the overlap a triangle with a base of 6 and a height of 3 this area we found to be 9 square cm as well. Mike McCollum ____________________ I though of this problem a little differently than just two squares. This problem could just as easily have said "make a 90 degree angle from the center of a square with 6 cm long sides that encloses the most part of the square" I tried doing this many different ways, making another square, making a 45,45,90 triangle but all of them had the same area. Therefore maximum area would be 9 cm squared. Bipin Mujumdar ____________________ The largest area possible for the overlap would be 9 square units. When rotating the square around its center all overlaps yields a 9square unit area. When the base of the second square is perpendicular to the ride side of the other the area is 3*3 or nine. When you rotate it from that position by 45 degrees it is nine again because after drawing the median of the overlap the area = .5*3*6 which gives an area of nine again. To measure the overlap at odd locations I graphed it on .5 inch graph paper. Again in all locations the area was 9 square units. Since every value was 9 no matter what, the largest area was nine square units. Gino Perrotte ____________________ Since the squares are 6cm x 6cm, I broke each square into 12cm smaller square units. The center of each square is located at the intersection of the lines of the third block down from the top of the square, and the third block from the side of the square. Therefore, place the vertex of one square at that intersection of the other. It doesn't matter which square is used for the vertex and which one is used for the center. Likewise, it doesn't matter which vertex is used on the square that is used. When the squares overlap you will see an area of 9 blocks, or 9cm^2. Matt Bouton ____________________ The most that the overlap area can be is 9 square centimeters, as demonstrated by the three possible rotations of the overlap area, shown in my drawing below. Nate Yarbrough ____________________ The largest possible overlap area is 9 square centimeters. See drawing below. Richard Bryer ____________________ The largest possible area would be 25% or 1/4. The area is 36, so 1/4 would be 9 centimeters squared. Even if you rotate the square, keeping the vertex on the center, the largest area is still 1/4 or 9 centimeters squared. Drawing is shown below. Vashti Bandy, Donald Fischer, Susan Quan, Alex Mikuliak ____________________ These students have submitted solutions stating the maximum overlap area is 9 sq, cm. They each argue that the maximum overlap is 1/4 of the fixed square. This problem is a good one to investigate using the Geometer's Sketchpad. If you have Sketchpad and your Web browser is properly configured to open sketches, you can view these three illustrations. Previous page || Next problem || Previous problem || Table of Contents || Forum Home Page