I’ve written previously about why I like the books from Art of Problem Solving:
Why I like the Art of Problem Solving Books
and I got another nice reason to like them today with an excellent problem in our geometry book’s section about parallelograms. I’ve written two other posts recently about parallelograms in the last week, too – we had quite a lot of fun in this section 🙂
Patrick Honner’s Challenging Parallelogram Problem
Cleaning up an Oversight in our talk about Patrick Honner’s Parallelogram Problem
This section of our Geometry book tilted a little more theoretical than some of the previous sections. This tilt led to some good conversations about techniques of proof and also equivalence. The chapter ends with a statement that these three basic properties of a quadrilateral are equivalent:
(1) The opposite sides are equal,
(2) The opposite angles are equal, and
(3) The diagonals bisect each other.
If any one of the above statements is true, the quadrilateral is a parallelogram. It is pretty tough work to get through all of the examples and proofs that lead to this statement.
The exercises at the end of this section end with a real gem of a challenge problem:
Problem 8.3.7 The diagonals of a convex quadrilateral ABCD meet at E. Prove that the centers of the circumcircles of triangles ABE, BCE, CDE, and DAE are the vertices of a parallelogram.
This problem seems pretty intimidating when you first read it. In fact, when I read the problem for the first time it was certainly not obvious to me why the statement would be true. What do the centers of those circumcircles have to do with each other?
We started with a simple picture
and then discussed how you would find the centers of the circumscribed circles. My son remembered that the center of the circumscribed circle of a triangle is the intersection of the perpendicular bisectors of the sides. This observation led to the following picture:
You get the idea from the picture that you may indeed have a parallelogram (though, less so from the picture we had on the whiteboard – ha!). Even with the idea that you may have a parallelogram, there’s no doubt that this new picture is pretty intimidating.
But, intimidating or not, it led to two really great discussions. In the morning we focused on definition (2) above and found why the opposite angles in quadrilateral XYZW above are equal. Along the way we learned about the angles made by pairs of intersecting perpendicular bisectors in a triangle.
In the evening we came at the problem a different way by understanding why the opposite sides of quadrilateral XYZW are parallel.
So, I was happy to have the opportunity to talk through this problem with my son. I think it is really important to learn how to approach problems that seem intimidating (or problems where you have no idea at all where to start). I had no idea how to solve this problem when I first read it and I really hope that our two talks through this problem helped my son see how to make progress in this type of situation.
The math lesson I’m trying to get through to my son is one that Julie Rehmeyer talks about in this interview:
Julie Rehmeyer’s “Inspired by Math” interview
What’s always stuck with me from this interview is the story that begins around 31:30 and in particular the part beginning around 34:40 about proving that 0 + 0 = 0.
She comes to meet her adviser, Phil, and is stuck on problem about proving that 0 + 0 = 0. Phil has not worked with the axiomatic system that Julie is studying and does not immediately know how the proof will go. He tells her, though, that he knows that if sits down and thinks about it for a bit that he will figure it out. He knows this because he is a mathematician and he’s learned how to work through proofs like this one. He tells her that if she sticks with math for a while that she’ll develop that same set of skills, too.
That’s a better summary than I could ever give about what I think working through the Art of Problem Solving books helps me do for my kids.
Mike's Math Page 