Thursday, July 30, 2015

Sometimes one direction is a lot tricker than the other

I was looking at a twitter post yesterday:

Which I will paraphrase below:  Given a square ABCD with an embedded equilateral triangle
ADE find the measure of angle CEB.

First: there was some disagreement to how easy this problem would be. I threw this one over to my beta tester who can angle chase but hasn't yet covered the Pythagorean theorem. I had included some notation to show the congruent sides in the triangle had started to include the square but then erased that on second thought. Unfortunately, it could still be faintly seen which ruins a fair test but given that much information this did not present much problem. His thinking went as follows: mark the 60 angles in the triangle. Then note the 30 degree angles at CDE and BAE.  There was a muttered "I need something else" followed by "Oh there are isosceles triangles".  With that he marked in DCE and DEC as 75 degree angles and did the rest of the math to find CEB is 360 - (75 + 75 + 60) = 150. This process lasted probably less than 2 minutes. All of which points out that this problem is a fairly standard angle chasing sample and you can provide some vague strategies like identify all the congruent sides and what kind of triangles are present that will help to unblock kids.

However, I had seen this same problem earlier this summer in its inverse form. Given the figure above if the triangle CEB is a 15 -15 - 150  isosceles triangle  (or two 15-75-90 right ones) prove the lower triangle is equilateral.  Interestingly stated this way the problem turns out to be a great deal harder. The key to the prior version was to find the isosceles triangle via its congruent sides and then calculate the two 75 degree angles. In this version you essentially want to go the other way. Prove both angles are 75 degrees so you can show the sides are also congruent. 

After a lot of experimentation over several hours I arrived at the following construction below:

By angle chasing you can easily find DCE is 75 degrees. So it remains to find CED. This will be done via reflecting the triangle CDE. First extend the line through F and G and then construct a line a 60 degree angle at HCG.  All together HCE is thus 75 degrees which will be our center angle when finding congruence.  This also gives you a new 30-60-90 triangle CGH. Because of that CG is half the length of CH. CG is also half the length of CB and CD because its in a square. So CH = CD. Then triangle CHE is congruent to CDE via SAS.  Based on that CEG = CED = 75 and you're done with some mechanical followup.

Same problem but this form would definitely stump most kids on first glance. Which makes this sort of like the geometric equivalent of a cryptographic hash function.

In which I find another tricky variant.

Looking around I saw this problem on gogeometry:
Initially this looks completely different than what was above. But pay attention to the right angle and congruent lines. If you reflect ABC to make a complete square and draw in BD you get this:

Which is just the equilateral triangle in a square with the 15-15-150s on the bottom rotated on its side a bit.  This version is just as tricky as above: see: for the key to finding the solution given these different unknowns. (Its a shame there isn't a geometric version of if I can find a single construction that matches the constraints, then that must be the answer)


Bringing this back to the realm of what might be usable for my math club:

1. I really like the idea of having kids try the problem both ways to see the difference. I'd favor doing it the hard way first for a few minutes.
2. Random: I should look more into cryptographic hashes.  This would be really cool as a bookend if we could do some simple hashing and exchange messages as our main activity.
3. It would be interesting to investigate the relationship between the 2 triangles within a square as you vary the angle sizes. Offhand  I'm think this is inverse tan x and inverse tan 1 -x.

Addendum: A simpler variant on this theme: part2

Monday, July 27, 2015

The Geometric Mean

I've been thinking a bit about the geometric mean this week after it turned up in two separate 
problems I've been looking at. In the first which is fairly simple given 2 circles that are tangent to each other the segment from the their intersection to the tangents on the sides is equal to the geometric mean of their radii i.e. $\overline{CD} = \sqrt{\overline{CB} * \overline{AC}}$.  Hint: it helps that triangle ADB is a right triangle.

In the second there is a triangle with an altitude that has two more perpendicular segments drawn from the other sides to it (point D) and then 3 inscribed circles are added in.  Again the middle circle's radius is the geometric mean of the other two.  $\overline{MN} = \sqrt{\overline{KL} * \overline{HI}}$. This figure has a lot of similar triangles.

I can prove both problems separately but it seems like they are connected and I don't yet see how to do so.

Further Riff has a discussion of the arithmetic - geometric mean inequality theorem I happened to read this morning.  Its very handy for solving the following: $(a+b)(b+c)(a+c) \geq 8abc$

Monday, July 20, 2015

Voronoi Diagrams

After a trip to the Field Museum in Chicago I've become interested in whether voronoi diagrams would make a fun exercise topic for the club.

So far I'm not sure if they are accessible or not. Philosophically I haven't really touched topics where the actual math is beyond the kids. I know that watching videos on numberphile even for more advanced topics that can only be summarized and described still interests my own kids. So a day organized around these diagrams might work despite the lack of theoretical underpinnings.

Looking on the internet, I've found the following activity which looks cool but is aimed at High School:  The reliance on circumscribing triangles
and perpendicular bisectors are probably blockers.

I could also probably walk everyone through the construction process which is kind of a neat art project.  f I had computers we could use several different programs to generate these as well. This is also potentially an extension of a tiling exercise, something I hope to touch next year. Finally in theory, I could have everyone come up with approaches to efficiently dividing an area around a set of  points  over half the time and then guide them through the process in the back half especially with a set of protractors.

Tuesday, July 7, 2015

Summer Reading

This book just arrived in the mail yesterday. I tried the first few problems out last night to escape the heat and they were pretty fun.  They also despite the title look potentially adaptable for math club subject to how many fourth vs. fifth graders we have.  Oh and I broke the 2000 hits mark this morning for this blog. As I joke with my wife, I'm definitely writing on a topic of specialized and limited interest.

Wednesday, July 1, 2015

How to make homework work

Just yesterday AoPS released a new online problem resource for Middle School level (MathCounts) material: I took a quick look last night and it seems fairly well done. Not too many frills, beyond a leader board but it does do full answers not multiple choice and the problem set is appropriate. This would make good drilling material for those interested in improving on contests like Knights of Pi as well.  Unfortunately, my beta tester is in the midwest with his grandparents. Perhaps when he gets back I can bribe him with enough m&m's to try it out for 10 minutes and give me his impressions despite it being summer break. If it does pass muster I'll point kids this way when we get close to the various contests it would help with.

This brings me back to the topic of homework. As I've mentioned before I did not have much success giving a challenge problem a week out to the math club last fall. So I adjusted and focused mainly on in class activities for the rest of the year. Quite possibly I'm setting myself up for repeat failure but I'm planning to give it another try this year. I don't want to do anything rote (there's plenty of that in regular school assignments). However, motivated by some posts around the blogosphere I do think these type exercises are a particularly good way to build problem solving skills. The key is getting the kids to even try them,  So my tentative idea is to do a reward system. I'll give a point for every sheet submitted showing some work done on the problem (maybe 2 points for a correct solution). We'll track it on a bar graph that everyone can see. Then when the group reaches a threshold I'll bring in some kind of treat. This mostly  mirrors what the various teachers do to reward good behavior so I'm hoping it will have some efficacy.


Here's my first attempts doing geometry drawings with geogebra: