A Hoop For Jumping From Middle School to Being a Teacher
Consider this problem from Ontario's Middle School Examination For Entrance Into The Normal Schools.....July, 1919.
Problem 4a: Trisect a given parallelogram by two straight lines drawn through one of the angular points.
Note 1: You tools are an unmarked straightedge and a compass, i.e. the 1919 equivalent to Geometers' Sketch Pad.
Note 2: You can assume that it is the parallelogram's area that is being trisected, as trisecting the angle itself is a "little difficult."
Hint: If you connect opposite vertices of the parallelogram with a line, you form two congruent triangles with the same area. Now, your new but simpler problem: how can you draw one line through the vertex of a triangle to its opposite side and divide the triangle's area into the ratio 2:1?
Solution Commentary: To get you started, suppose your parallelogram is labeled ABCD. Then, trisect side BC with points E and F. Draw AF. Why is the area of triangle AFB equal to 1/3 the area of the parallelogram ABCD?
Now, you need only repeat the process using side CD.
But, wait....you will have to first refresh your construction skills....how do you trisect a line segment?
