Wednesday, September 20, 2017

Teasers for this year

Things are moving along. I have a provisional process to get going in 3 weeks and I've send out the initial signup forms.

Some of the challenges so far:

  • Pickup policy.  Without any backup I'm having parents give permission for everyone to self-release.
  • Fees.  Its harder to collect money under the new structure. I still have to check on the reimbursement process as well.
  • Gender Balance - this is looking promising so far this year. I'll see when the actual signups start coming in. 

Here are my teaser problems for this year.

Inline image 1

There are 100 people in line to board a plane with 100 seats. The first person has lost their boarding pass, so they take a random seat. Everyone that follows takes their assigned seat if it's available, but otherwise takes a random unoccupied seat. What is the probability the last passenger ends up in their assigned seat?

Inline image 3

Friday, September 1, 2017

Status Updates


School is about to start next week and I am still trying to get all the logistics in place for this year. Hopefully by next week I'll have made contact with the ASB coordinator and have an idea when things will start, how to advertise etc.  I'm definitely getting excited/antsy and am ready to interact with kids again.

[Update] - I put out a FB post on the school page and have already generated 11 inquiries of interest including several girls. So far so good.

To Do:

  • payment process
  • pickup procedure
  • Remember to offer to email students directly now.


My new revised repository of interesting internet problems is here:  The new version has my custom note button, some rough levelling  and proper attributions.

Of Note:

Found on FB, original source unknown. I like this one as a starter. [With the numbers switched up a bit]

Tuesday, August 15, 2017

Elegant Solutions vs. Hacking

This is a study in contrasts around a fun problem by @eylem:

An elegant solution to this would be as follows:

  • After angle chasing to find angle DEA is 45 degrees extending DE to make a full right isosceles triangle with side lengths of 15 sqrt(2)/2. 
  • Use the Pythagorean Theorem to find the missing side length of AGE. This gives you the base of the triangle DE also,
  • Then note AGE is congruent to the halves of CDEs so you also know the altitude.
  • Apply the triangle area formula.

But there are other shadier options for solving the problem:

Instead, we can take advantage of patterns in the values of the lengths and the then use knowledge about 3-4-5 triangles (see: to crack the problem.

This actually exposes a bit of structure that was not seen in the first solution.  If you didn't think to try the educated initial guess, you could more directly find it by setting up a simple quadratic equation based on the Pythagorean Theorem:

$$ (5\sqrt{5})^2 + (15 - x)^2 = x^2 $$

Tuesday, August 8, 2017

How I use Twitter

This post started with some musing about the meta conversations occurring online right now in twitter over hashtags.  There was enough activity that I was distracted for a bit and then I came to the realization: This is not why I'm on social media.

I started logging on to Twitter in January of 2015 after seeing some math posts on google+ that seemed to indicate there was a lot of interesting activity going on.  After a few weeks I was convinced:  twitter was a great place to find other people actively discussing mathematics and working with kids.

I started by monitoring #mathchat which has a mixture of spam/ads and real people.  From there I found #mtbos which is much more focused on teaching, has less spam and a more communal feel. Over time, I've been slowly growing my follow list of people who post interesting ideas.  Building this network has tended to focus the tweets and made the experience more useful.  Although compared to average I have both less followers and follow less people which probably reflects my own style for engaging with social media.

What I'm finding over time is there are several different types of posters that I enjoy the most.

1. Puzzle and problem producers. These are folks like @gogeometry, @eylem, @cuttheknot, @five_triangles, @sansu_seijin, @solvemymaths.   They regularly post interesting problems that I like to try out and reuse.

2. Animated Gif Makers. There are a ton of really interesting animated math gifs being produced by folks like @gohio, @dynamic_math, @beesandbombs.

3. Individual Bloggers. These are folks who's blog or channel I regularly read anyway like @mikeandallie, @hpicciotti, @fawnpnguyen, @mrhonnor, @math8_teacher, @standupmaths
I've found a lot of sources for ideas for ideas about activities or pedagogy this way.

4. Conversationalists: These folks are usually less for blogs and more for the conversations they produce in twitter itself. @mpershan, @trianglemancsd

The flip side is that twitter is not all goodness. Its very hard to express complex or nuanced ideas in the limits of a tweet. And then there are viewpoints out there with which I don't agree. I'm definitely susceptible to the "But someone's wrong on the Internet!" phenomena, especially if a person posts a lot.  So I try very hard to filter these out rather than responding most of the time. Arguments in 140 character tweets are not terribly exciting. In fact, there are even a few folks I stopped following even though I agreed with them because most of their time was spent in debates that were not useful for me. In the end, my mantra is try to produce the content that I would also like to consume.

Monday, July 31, 2017

Angle bisector

I've officially reached the point of the Summer where I'm missing interacting with kids besides my own.  In the meantime, this is another geometry walk-through of a problem from Cut The Knot this time. It's a  good example of trying different ideas to utilize angle bisectors.  Plus, if you haven't checked out this site, its well worth the time.

Easy Observations

To start off as always I took a look at the drawing and went with my instincts: that really looks like AZ is bisecting BAD and is 30 degrees.  I was so sure, I didn't even try modelling to confirm that impression before moving on.

The second observation I made was that BI was also the angle bisector of angle B since it intersected the other 2 bisectors at point I (the incenter).

Next I went through the obvious angle chasing. Initially I assigned a to the 2 bisectors of angle B and b to the 2 bisectors of angle C. So 2a + 2b = 60.  After walking through this for a while, I realized that it was a lot more convenient to let b = 30 - a and only use one variable.  The one interesting thing that popped out was EIB was 30 degrees. What I really wanted next was angle CDE or BZD and those were not deducible yet.

Initial Targets

From there I noticed if DAZ was 30 that AEZI would be a cyclic quadrilateral. So I needed to somehow show that was the case. Nothing immediately came to though for finding more angles or side lengths to do that.

I like to walk backwards from the angles sometimes to find more patterns. This showed CED was also 30 degrees and triangle EIZ was isosceles. If I could get to that point I would also be closer. But yet again I didn't see a connection to make.

Some Experiments

I picked this up again a day later, and tried a couple of experiments with the angle bisector. First, I thought it might be easier to manipulate if I extended the triangle to make it isosceles AD would extend to be a perpendicular bisector.

This produced some more semi-interesting angle in the new portion but no similar triangles or such and didn't seem helpful.

Then I decided to instead increase symmetry by mirroring on the right with an explicit angle bisector:

My hope was that I could prove the right and left similar and thus show the left was also a bisector.  It was easy starting with the bisector to show the right hand side was  a cyclic quad and it was noticeable that the combined isosceles triangle would form a larger cyclic quad but again I didn't quite see how to connect all of this. It also needed to be shown that the bisector on the right really intersected the other lines at the same point.

I then thought about creating multiple equilateral triangles through extension. This seemed interesting since there would then be a giant parallelogram. Again, it was interesting but there was not much I could do with this. I added the diagonals in which intersect at right angles but even that didn't give me new ways to solve the problem.


I stopped experimenting but then while trying to go to sleep I thought of a new angle.  Point Z was also the incenter of the smaller triangle ABD if I could prove that DE was an angle bisector I would have a way forward. That was attractive because I could use the angle bisector theorem. I just needed to show  BD and AD were in the same ratio as AC and BC.

Since those two segments weren't adjacent I trying picturing ways to move them closer. By overlaying them or adding reflections. Eventually I settled on reflecting AD over to intersect AC and immediately I got excited. When that happened I formed a new equilateral triangle and a parallel line (DF) and similar triangles (ABC and CDF).  From there I could pretty much see the way forward:

Final workup as transcribed by @CutTheKnotMath (Which is a lot prettier than my sketch)
Draw DFAB. By the Exterior Angle TheoremAFD=FDC+FCD=ABC+FCD=60, making ΔADF equilateral. In particular, AF=AD.
A Problem from the 1985 Balkan Mathematical Olympiad (Shortlist), solution 2
By the Internal Angle Bisector theorem, AEBE=ACBC.
By the Thales' theoremBDBC=AFAC=ADAC, i.e., ADBD=ACBC, implying ADBD=AEBE.
By the converse of the Internal Angle Theorem, DE is an angle bisector in ΔABD and so is AZ. Thus DAZ=30.

So this took in all 3 separate 20 minute sessions over 2-3 days with several dead ends as is often the case for me with trickier problems.  What also seems to be the true is that experiments tend to deepen my understanding of the construction. So while they aren't always fruitful they lead the way towards the eventual productive ideas.

Trigonometric Technique

What's also fascinating is looking on the site at the other solution. First it would not have occurred to me to use Stewart's  theorem.  So an interesting technique to remember for future use. But also new to me was the variant here:

$$AD = \frac{2bc \cos(\frac{A}{2})}{b +c}$$

I found a good explanation here: Basically if you use the law of cosines to find the area of the subtriangles ABD and ACD as well as ABC and simplify this form pops out. If you think to go this way, it avoids the need for auxiliary lines (as often happens when trig is introduced).

I'm particularly proud of this effort because my solution was added to the site. Now I just need to start working so that I can tackle some of the hairier inequalities problems discussed there.