A quick post here – I want to share something delightful that a few Geometry students did this morning. We had our last test of the winter term today and here is one of the last questions:

Prove that the points A (x, y), B ( x + 1, y + 3), C (x + 4, y + 5), and  D (x + 3, y + 2) are the vertices of the parallelogram ABCD. Prove this is true by one of the following two methods:

  • By showing that one pair of opposite sides are congruent and parallel.
  • By showing that both pairs of opposite sides are parallel to each other.

So, I was hoping that the majority of my students would take the quick and easy option of calculating slopes rather than messing with distances. I also hoped that the coordinates having variables in them would make them slow down, be careful, and remember a touch of algebra. I grade page by page and I have graded three of the papers with this problem on it. One student said ‘We can let x and y be 0 so the coordinates are (0,0), (1,3), (4,5), and (3,2)’ I love this thinking. She avoided the worry of dealing with the variables here. It’s a little slippery to determine just how clearly she was thinking here. She might have just been dodging a bullet. One student said ‘I will first transform this parallelogram by the vector <-x, -y> and then we will have the coordinates A’ (0,0), B’ (1,3), C’ (4,5), and D’ (3,2)’ Now, it is ABUNDANTLY clear that he knew exactly what he was doing. I’m so delighted by this that I felt I should share.

This and my great AP Stats classes today made for a pretty terrific day!

Different Perspectives

A quick reflection here before I wake up my kiddos.

Yesterday in BC Calculus we had onto of our weekly problem days where we (mostly) put aside our current Calculus work and look at interesting problems that may or may not involve any Calculus at all. Here is problem #1 from yesterday (a problem I borrowed from @bretbenesh.

A mountain climber is about to climb a mountain. She starts at 8 am and reaches the summit at noon. She sleeps at the top of the mountain that night. The next morning, she leaves the summit at 8 am and descends using the same route she did the day before, reaching the bottom at noon. Why do you know that there is a time between 8 am and noon at which she was at exactly the same spot on the mountain on both days? We should not assume anything about her speed on either leg of the trip.


One of the things I enjoy most in teaching is seeing/hearing different ways of attacking a problem. When I read this problem I immediately sketched a height v time graph with the base of the mountain and 8 AM as the origin and the top of the mountain, noon as some arbitrary point in the first quadrant. A wiggly sketch connected the points. Day two has a y-intercept of top of mountain, 8 AM and an x-intercept of bottom of mountain, noon. No matter how I connect these points the sketch intersects my other sketch and I see the reason why. I’m surprised by this discovery, but I see it. In each of my 2 BC classes yesterday there was one student who saw through this problem and explained it away so quickly that I was wowed. In each case the student immediately switched to thinking of 2 people rather than one. If one person starts at the top at 8 am and walks down while the other starts at the bottom and walks up then they must pass each other at some point! Simple, clean, elegant. It’s fun to learn from your students, isn’t it?