The problem was from AMC8 and I original saw it referenced on mikesmathpage.wordpress.com

Points ABCD are the midpoints of a square and form a smaller square. If the larger square has area 60, what does the smaller one have?This one can be determined visually if you notice the four squares the larger one is divided into and the 8 triangle that these are further subdivided by. That means half the triangles are inside the inner square and half are out so the area is 1/2 of the whole. A lot of kids found the solution in variants on this observation. A few computed the side length of the larger square and then the area of the triangles which is a bit more work. Better yet when I asked what everyone had observed when trying this out with other quadrilaterals many of the kids had experimented and found the basics of Varignon's Theorem i.e. the inner shape is always a parallelogram and has half the area. I definitely want to do more investigations during further club meetings along these lines.

http://mrhonner.com/archives/17154 has a good proof that could be developed in class.

We then jumped into our first MOEMS Olympiad for the year. In my tryout beforehand I noticed a few things. First, by coincidence it has a simple decoding problem in the middle which meshed really well with the examples we had done that last few weeks. Secondly the final problem, arrange the 8 integers 1..8 around a grid (no center) with some specified sums for the rows and columns and then find the sum of the numbers in the four corners looked like it would take the most time. It was solvable algebraically but as predicted everyone who was successful ended up using a guess and check strategy. Many more kids this year took the full ~30 minutes to do the problems than last year which is probably a function of age i.e. fourth graders vs. fifth graders. Incidentally, the cool part of having some returning students is the ability to see how much they've grown in a year via their performance on these contests.

Secondly on review I still have kids who given a simple sum will plow through it from left to right despite me trying to stress looking for shortcuts via rearranging the numbers. As an aside, with problems like this I'm taking more advantage of class surveys. So I'll start by just asking for a show of hands of kids who added the numbers in order and then ask for anyone who did something differently to bring out the alternative strategies a little more efficiently. I should definitely stress this point more and do some examples of regrouping in club.

My favorite part of the whole experience by far was the wrap up when I had the kids show their solutions to each other. All our work on listening seems to be paying off. Despite there being 5 problems rather than our normal one, everyone was rapt during the demonstrations and I had lots of volunteers wanting to show how they had found the answers.

Also new this year, I planned for kids finishing early and brought another exercise from "This is not a Maths Book". This time we did the page on Pascal's Triangle which involved building your own triangle and colouring in evens and odds to search for patterns. This proved really fascinating for the 6 kids who were done early and kept them busy (and non-distracting) to the others who were still finishing. Again this year, I plan to do a full unit around Pascal's Triangle some time in the winter. There's too much good stuff here to not touch again.

Finally, I'm trying a problem from a new source http://www.cemc.uwaterloo.ca/resources/potw/2015-16/POTWD-15-NA-09-P.pdf for the take home problem. I'm hoping again that at least some of the kids will notice that 1001 is a common factor in all numbers formed this way. We'll see what they come up with when everyone comes back from Thanksgiving break.

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