A former colleague of mine, Lisa Winer (@Lisaqt314) tweeted a problem at me last night. Wednesday night I s my basketball night and then I curled up to watch some Netflix with my wife, so I did not see the problem until this morning. It has since been making the rounds a bit. It is called ‘The Hardest Easy Geometry Problem’ and you can find it at https://www.duckware.com/tech/worldshardesteasygeometryproblem.html
I started working on it on my side board today and it caught the attention of my BC students. One of them found a solution using trigonometry and I constructed the triangle in GeoGebra to confirm that he is correct. However, I still have not found a way to solve it geometrically. I reduced the problem to four equations with four unknowns but the matrix is singular and I could get no solution. I did, however, have fun playing with it and watching a number of my students dig in.
In Calculus BC today we talked ourselves into the area formula for regions bounded by polar curves and we had great conversations about it in both of my sections. In each class I had at least one student remember some area formulas for triangles that are rarely used and that help serve as the basis for the integral involved. I was pleased with each of those classes today.
In geometry we are working with quadrilaterals and a recent HW problem presented the students with a parallelogram and some algebraic expressions to deal with. Most of my students made an assumption regarding the intersection of diagonals for the parallelogram. They correctly assumed that they bisected each other. I was pleased that they made this assumption but I made sure that they felt comfortable with an argument supporting that fact and then a series of questions erupted that carried us through the end of the day. Do diagonals bisect each other for all quadrilaterals or just parallelograms. A couple of quick sketches at their desks implied that it was not always true. A quick visit to GeoGebra seemed to convince them. Then a student asked if a quadrilateral could have congruent diagonals if they do not bisect each other. A few more sketches and then the guesses started flying in. It did not take long to guess that an isosceles trapezoid would fit this bill. Again GeoGebra confirmed our guess. What next? How about the triangles for,Ed when the diagonals cross? Are they all congruent? Are they congruent pairs even? Quick feelings that the ‘side triangles’ are congruent but the top and bottom ones are not. Right again! But my favorite part came next. I did not plan on talking about area for a couple of days still, but the moment felt right. I asked if we could deduce an area formula for this trapezoid. Now, last year at this point I had a student suggest drawing one diagonal to find two triangles. Standard and clean. I also had a student suggest dropping two altitudes from the ends of the shorter base. Again, a nice standard solution. I had one student suggest rotating the trapezoid 180 degrees to create a parallelogram twice the size of the trapezoid. Not standard at all, but also kind of confusing for his classmates. This year, I had a student named James make a suggestion I had not head before. He asked me to draw segments from the end of the shorter (upper) base down to the midpoint of the lower base. This created three triangles all with the same height. I took a picture of the sketch we Made on the board. That is the photo on top of this post. I must say that I am completely delighted at this clean and clear way of looking at this area problem.
A pretty good day overall, I must say.