Monday, December 26, 2016

Technology

I saw a funny ignite talk "Algebra Inferno" the other day comparing disliked teaching practices to the various circles of hell a la Dante.



Among the sinners list were the survivalists: those who refused to ever allow the use of technology in their classes.  While I chuckled at the clever pun, I don't really totally agree with this point. My contrarian instinct is that there is a huge body of math out there that doesn't need calculators that one need never stray from and quite successfully conduct a lesson. And on the converse most of the math lessons that try to use them often turn out to be more about practice punching buttons than thinking mathematically.

There's some larger existential questions wrapped in this debate. In the era of  ubiquitous cellphones and Wolfram Alpha what parts of elementary school mathematics are relevant and can you skip them without throwing away the ladder to higher level skills?

Leaving that question aside, in practice, as often is the case I'm more pragmatic than my initial instincts. The other day my son was working on a problem and asked if he could use Desmos to plot the graphs of some inequalities. That seemed like a very reasonable usage and I'm happy how much he is excited by desmos having only been recently introduced to it. So feeling like another hand plot wasn't needed, I said "of course."

I don't have that  luxury during Math Club where there are neither calculators or computers around. But were that the case these are just a  few of the uses I  think really are worthwhile:

  • Investigating the patterns of digits in repeating decimal numbers is vastly sped up by just trying them out.
  • In the age of infinite precision calculators its now possible to check all those modular arithmetic stumpers  like which is bigger 63^45 or 33^54  directly in python.
  • I love the use of 2-d and 3-d graphs as long as they are a natural extension of a larger problem.
  • Geogebra makes a nicer version of pen and compass constructions and is very useful in exploring more complex geometry proofs.
  • We were recently doing a comparison problem at home between 2^1/2, 3^1/3 and 5^1/5.  Visualizing the graph of x^1/x in desmos made this much richer. 
  • If you want to practice finding factors for larger numbers like say 2017, calculators help speed things up quite a bit.

Tuesday, December 20, 2016

Virtual Math Club for Winter



Here's the cipher I sent of for the kids to decode while they are on break. I'm offering double homework points towards candy for solutions so hopefully some will give it a shot.

"oifi'y r trgo dcit rme ucji gc chhkdl lckf gzti xiscfi trgo hpkx ygrfgy kd rbrzm. z'ni imhceie zg wzgo r ykxygzgkgzcm hzdoif gc trji gozmby tcfi skm:

z
wfcgi
r dcit
cm r drbi
xkg goim irho pzmi bfiw
gc goi goi wcfe ykt cs goi dfinzcky gwc
kmgzp z ygrfgie gc wcffl rxckg rpp gocyi wcfey hctzmb wzgo ykho sfiakimhl
xihrkyi ry lck hrm yii, zg hrm xi iryl gc fkm ckg cs ydrhi woim r dcit bigy rpp zxcmrhhz yiakimhl."

Monday, December 19, 2016

Winter break cyclic 3-4-5


This is another exercise in documenting geometry problem solving. I chose this problem because again it has a 3-4-5 triangle within it and the overall setup is very simple.  It also shows how one can circle around the correct solution,  as it were, before coming to it.  


Given cyclic quadrilateral ABCD where  ABC is a right triangle and AB = BC = 5  and 
the diagonal BD = 7 find the area of ADC.



Thought process


1. Angle chasing:  \(\triangle{ABC}\) is right isosceleses. Since  ABCD is a cyclic quadrilateral, that means we can angle chase a bunch i.e. \(\angle{BDC} = \angle{BAC} =45^\circ\).
2. Cyclic also means that the product of diagonal portions is equal i.e. \(AE \cdot EC = DE \cdot BE\)
3. Also there are a bunch of similar triangles: \( \triangle{AED} \sim \triangle{BCE} \) and \( \triangle{CDE} \sim \triangle{ABE} \)
4. Realize 2 and 3 are the same basically saying the same thing.
5. At this point I had a general idea that I'd need to use the length of both diagonals as well as  the similarity of the triangles to find the area.
5. I started the algebra with the Pythagorean theorem and similar triangles to find the lengths of AE and ED.  This looked fairly messy even going in although generally solvable.
6. So I stopped and checked with the angle bisector theorem around \( \angle{ADC} \) and realized that produced nothing new.
7.  I already strongly suspected just on visual inspection that the missing part was a 3-4-5 triangle. So once again I paused and checked in geogebra that my hunch was correct. (it was).
8. Then I started grinding through my first approach:  let a = BE and b = CE then:
\( \frac{BE}{EC} = \frac{AE}{ED}\) or \( \frac{a}{b}=\frac{5 \sqrt{2} - b}{7-a} \)

Which simplifies to \( 7a - a^2 = 5\sqrt{2}b - b^2 \)
In addition using the Pythagorean theorem: \( \overline{AD}^2 + \overline{CD}^2 = 50 \)
Then applying triangle proportionality \( \overline{AD} = \frac{5}{a} * (5\sqrt(2) - b)) \) and  \( \overline{CD} = \frac{5}{a} * b \)
When combined you finally end up with:
$$ 50 = \frac{25}{a^2}((5\sqrt(2) - b)^2 + b^2)$$
$$ 2a^2 = 50 - 10\sqrt{2}b  + 2b^2$$
$$ a^2 = 25  - (5\sqrt{2}b - b^2)$$
$$ a^2 = 25 - 7a + a^2$$
$$ a = \frac{25}{7}$$

9. I started to  solve for b after which point I could find AD and CD. But this was even messier looking than above and was heading towards an ugly quadratic equation.

$$ \frac{25 \cdot 24}{49}  = (5\sqrt{2} -b) \cdot b $$

10. That looked super messy but I realized what I really wanted was \( 1/2 * AD * DC  \) which
comes out to \(  1/2 \cdot \frac{25}{a^2} (5\sqrt{2} - b) \cdot b \) and you can substitute with the two derived results above to find the answer.

11. I was dissatisfied with the opaqueness and algebra of this method plus it never showed the 3-4-5 clearly so I started from scratch.
12. This time I played with the \(\angle{ADC}\). First I stated thinking about breaking up the 3-4-5 into a square and the various 1:2 and 1:3 triangles. But then I realized that I had two 45 degree angles that  could be extended and we know the lengths of the sides of those triangle  and from there the drawing below immediately fell out.  Framed this way the problem was much simpler.


*

Friday, December 16, 2016

AMC 8 Results

The wait is finally over. We received the results for the 2016 AMC 8.  Now comes my least favorite part sending them out to the parents.  This is the message I ended up with this year:


The results for the AMC 8 contest finally arrived.  First thanks for participating. I'm proud of how everyone did and this is just the beginning, hopefully, of AMC contests for everyone. I want to stress again also to treat this like a baseline on an above-level test. The contest is meant for Middle Schoolers. Since its nationally normed, keep these scores for future years. I'm hoping everyone will generally see growth over time. 
Some Resources:
https://www.artofproblemsolving.com/wiki/index.php/2016_AMC_8
This has all the problems and solutions for them. If your child is still interested, it can be valuable to go over the problems. Feel free to email me if you have any questions about the questions. 
MAA will eventually publish national statistics when they're done scoring all of the entries. These will be found here: https://amc-reg.maa.org/reports/generalreports.aspx  I don't
generally think these are too useful in our case since the data is normed for 6-8th grade.
 

Overall I'm very happy we participated.  Hopefully most of the kids will take this again  in future years. As an aside, the opportunity to participate is very site dependent here. I wish there was a more district wide policy so that all middle schools always offered the chance.

Moving forward I'm feeling like some encoded message fun over the break:
http://mymathclub.blogspot.com/2016/12/virtual-math-club-for-winter.html


[Update]: the overall statistics were just published. One thing  that stuck out at me was WA state is a bit of a math powerhouse although sadly not much of this is coming from our district. Overall the state had highest mean score, median score and was the only one where you needed a perfect score to be in the top 1%.



Thursday, December 15, 2016

12/13 End of the Quarter

And just like that, another quarter has wound down.  I'm always amazed how fast time flies. Its a lot of work planning and running the sessions of Math Club and yet I'm still always torn wishing I had more time with the kids.  This day was more crowded than usual. I had to fit the second MOEMS Olympiad in, go over the problem of the week and there was the end of the  quarter game day to run. Clearly, something had to give and I ended up planning to sacrifice part of the time I normally would allot to game day. I'm considering running another one on the first meeting in January after we go over all the house-keeping  to make up for cutting it short.

As of yesterday I also finally received the updated rosters. I sadly lost 2 kids to scheduling conflicts but I have one new one joining and the overall numbers are still at 13 which is a good size.  Almost every time a student leaves I still feel a little bad though. There was one enthusiastic boy in this group  who often wanted to mention something one on one whom I really think would have had fun if he could have continued and a girl I lost to a conflicting choir practice last year that I ending up thinking of while I looked over the roster.

The problem of the week was only lightly participated in and I'm going to need to invigorate it a bit. So again I had kids work on it on the board: sketching out numbers and connecting the factors to see how far they could get. Since I'm finally getting used to the room, I took full advantage of the whiteboards on both sides and had 4 kids working at once.  That works great and sometimes a few extra kids come over and form a group which I like watching. What I didn't do but I will in the future was explicitly hand out paper to have everyone else working at the tables. Otherwise, some number kids don't watch or start working on their own  and sadly don't volunteer that they are missing the paper needed for scratch work. This is one of the areas where I think you just need to bridge the gap. Once  I imagined the group of kids always having a notebook and diligently bringing it. Maybe that works in Middle School but in fifth grade you have to do more to keep everyone on track.

There were a few behaviors I noticed during the Olympiad that I'm hoping to keep working on.  The first was 2 or 3 kids got stuck on one of the problems and basically gave up and turned the contest in with time still remaining. In each case, I encouraged them to take advantage of the time and keep working on the problems but it was a hard sell. (I also had a larger group that didn't figure out the solution to everything but kept working the whole time possible.) Persistence is one of those key attributes that I'm really trying to emphasize. I'll probably talk about it again in January but I'm still thinking about what the best way to handle this is. My first idea is to ask some of those who kept going to talk about what their thinking/strategies were when stuck.  Despite my skepticism sometimes of the value of most of the growth mindset theory that's really what's at play here. How do you get kids to buy into continually thinking about a solution (which they may not arrive at) and not shutting down?  It's this uncomfortable space where I think the most learning occurs.

My favorite moment of the day actually happened right after the contest was done. There was one problem that involved figuring the missing numbers in a consecutive sequence of number given the sum of some of them.  Listening to two boys discuss it one of them said "I solved it using try and fail (guess and check)" and the other replied "I solved it with algebra."  One its always cool to examine the structure of a problem through different strategies. But  more importantly this represents the inflection point many of them are at between informal methods and algebraic thinking.   I find this transition to be fascinating. There comes a point when you see problems and you immediately model them using linear equations and formally solve. Often this comes at a trade-off where the previous conceptual / informal reasoning takes a back seat for a while.  I find this analogous to how standard algorithms usually supplant informal computation strategies after they are initially learned.

Finally, I brought pente, prime climb, trezetto and pentago as well as a deck of "24 cards". These were all great hits and the kids immediately settled in once they had finished the contest.


Wednesday, December 7, 2016

12/6 Catching Up

As I alluded to in my last post I had to miss last week's Math club meeting. This was one of those times when expanding to two instructors really paid off. Thanks to the other parent, I was able to have the kids do a joint session and didn't have to cancel the day.  This was also made a bit easier by the fact that we were scheduled to do the first MOEMs Olympiad. So for the most part the kids were working by themselves on the problems in a proctored setting.

Nevertheless, that meant  I had a lot of items to catch up on when I rejoined everyone this week. To start off I had all the kids talk about their recent experiences with the AMC 8 test and with the first Olympiad.  The general consensus was that the first Olympiad was pretty hard. This was our first time trying out the middle school division so I was unsure what to expect. I always do the contests myself before hand to gauge their difficulty and form ideas about what I expect the kids to have trouble with. I also thought this was going to present a high degree of challenge. Interestingly, I've now looked at the second in the series and its doesn't appear as hard. So I'm going to continue on with the experiment and not switch back to the elementary division for now.

Because they hadn't gone over the solutions as a group I decided we would do so now even though they had been handed out. My general observation is that most kids don't reflect on problems unless you setup the structure to encourage it (even with a solution set).  The drawback was that this was a week later and so there was less excitement than there would have been in the moment but I think it was critical to draw out the discussion and have everyone think more about the problems they hadn't been able to solve.  

I'm unable to directly discuss the problems but several observations I did have were:


  • Comprehension is a bit of a problem. For instance there was one problem that asked the kids to find only cubed numbers. Many of them totally misunderstood and just looked for all the numbers. I'm going to have a discussion about careful reading before the next version and see if that's enough otherwise we may need to do some group practice parsing.
  • Many of the kids are unfamiliar with cubes and exponents in general. That's not completely unexpected given the scope and sequence for the year and I don't usually pick problems that hinge on them. But I may need to make an exception to facilitate with the contest and plan a day around exponents to give the kids the tools they need. The key idea I'd like them to understand is how an exponent is just notation that can always be expanded out into multiplication.


By this point we had used up more than half of the time and I wanted to switch tacks and do some group work. So I went with the factoring puzzle I had previously found:


As I expected this was compelling and accessible.  As I worked my way around the groups, most of my discussion were around which numbers can you see most immediately i.e. the 5 sticks out first for most kids (and then the 2's) and how do you go forward from there.  I needed to ask a series of leading questions to get some of the kids to think about factoring and structure. I.e. since all these numbers share factors once you've factored one you're only missing one factor in any of the other ones. So it also makes a lot of sense to factor the easiest one 6160 first.  My favorite observation, was one students noticing that once you did have the factors and ordered them, the smallest factor belonged behind the largest value, the second smallest factor belonged behind the second largest value and so on.

By this point I ran a few minutes over schedule. That meant I ended up skipping past the previous problem of the week and the other items I had assembled for the day.  If only we could have kept going for another 30 minutes ... On the bright side, we're well setup for our normal routine next week which will be the last session for this quarter.

P.O.T.W
http://furthermaths.org.uk/docs/FMSP%20Problem%20Poster%203.pdf






Friday, December 2, 2016

Grab bag of problems



Sadly, I  have to miss Math club on Tue. given a family emergency. In the meantime, these are some of the problems I've enjoyed looking at recently that I might use sometime soon:








I've reached that point when I see something like this that I think: huh if I form a  right triangle using B to H as the hypotenuse, the remainder of one side is 3/5 r and the hypotenuse is r to the other side must be 4/5 r so its a 3-4-5 triangle.  There are a 1:3 and 1:2 triangle embedded on the edges of the circles as well.  To actually prove this is true just takes the Pythagorean theorem in my version but I suspect there are several other means to the same end. I'd probably scaffold this a bit.



A lovely factoring and deduction exercise. Not too hard beyond the large numbers.  Actually what might be fun is to adapt this and do it interactively in club.

You only need to find the ratio in one of the eight wedges (I drew in ABU) to find the total answer.
Also you could draw a different wedge by extending the inner triangle IJU but I find the first form a bit easier to reason about. Note: the 45-45-90 triangles and the other similar ones. The answer can be found through a combination of similar triangles and area addition starting from the 45-45-90.


The kids actually did this one. It was my second favorite to discuss. My first solution was a rather tedious exercise in the Pythagorean Theorem. There is a similar triangle version that is much simpler.


Some lovely divisibility rule practice embedded in this puzzle.


Find the smallest possible value of the expression:

$$\sqrt{a^2 + 9} + \sqrt{(b-a)^2 + 4} + \sqrt{(8 - b)^2 + 16}$$

This looks complex but ultimately boils down to the shortest distance between n points is a line! Unfortunately it needs a tiny bit of algebra to recognize the geometric representation and find the line so I'll have to save it for some time in the future.