Friday, April 20, 2018

4/16 Purple Comet Online Contest

Three years ago I discovered the Purple Comet contest @  It has close links to the AwesomeMath  and I really liked the problems in the old tests. So I tried it out with my fifth graders at the time.  That 2015 Experience discouraged me from doing it again.  Despite the kids theoretically having up to Math7 knowledge the contest was too hard and I needed material that was better levelled for them to be most productive.

Cut to this year when I have actual 6th and 8th graders and I decided to participate again.  My current motivation was less the problems themselves than the timing. We don't have any real contests to participate in during the Spring and I wanted to do one meaningful one for the kids who like doing them.  Since you have a testing window when you can administer the contest and it just needs a few computers, the overall experience is very low barrier (much easier than an AMC test).

Overall this year went much better. I split into two teams and we worked in the library. Both groups found answers to over half the problems and most kids stayed on target.  I had  few at the last 10 minutes who had reached their limit for the day which is not unexpected.

I'm once again trying to decide which problems to review as a group next week. I'm thinking we probably should only do 3 maximum so perhaps I'll bring a set and let the kids vote on which ones they want to see the most.

In the meantime and in no particular order here are my observations about a few problems I noticed and thought about while proctoring, now that its ok to discuss them:

Problem 17

Let a, b, c, and d be real numbers such that \( a^2 + b^2 + c^2 + d^2 = 3a + 8b + 24c + 37d = 2018. \). Evaluate 3b + 8c + 24d + 37a.

This one is phrased in such a way as to suggest non integer answers but just looking at it the temptation is to say what  if the two expression are completely identical and 3a = a^2, 8b = b^2 etc?  And if you compute the squares sure enough  \(3^2 + 8^2 + 24^2 + 37^2\) does equal 2018.

That's fairly easy for a problem at the end and its not a very satisfying method. So what I think they might have been aiming at was to subtract the two expressions and complete the square to get

\( (a-\frac{3}{2})^2 + (b-4)^2 + (c - 12)^2 + (d-\frac{37}{2})^2  = \frac{1009}{2} \)  The right hand side is 1/4 of 2018 so you can plug the original equations back in to get

\( 4( (a-\frac{3}{2})^2 + (b-4)^2 + (c - 12)^2 + (d-\frac{37}{2})^2 ) =   a^2 + b^2 + c^2 + d^2   \)

Then subtracting the right hand side again you get:

\(  4(a-\frac{3}{2})^2  - a^2 + 4(b-4)^2 - b^2 + 4(c - 12)^2  - c^2+ 4(d-\frac{37}{2})^2 - d^2   = 0 \)

Each one of those pieces is a difference of square for example the first one is \( (2a - 3 + a )(2a - 3 - a)  = (3a - 3)(a - 3) \) and if we set a to either 1 or 3 will zero out etc. You can do a similar operation with the other original expression and see that a = 3 for instance is the overlapping solution to both new equations. That gets to the original observation once you test the values in the original function. But it doesn't rule out alternate solutions where the pieces balance each other out in various ways.

Problem 16 

On  \( \triangle{ABC} \)  let D be a point on side AB, F be a point on side AC, and E be a point inside the triangle so that DE is parallel to AC and EF is parallel AB. Given that AF = 6, AC = 33, AD = 7, AB = 26, and the area of quadrilateral ADEF is 14, find the area of  \( \triangle{ABC} \)

A significant part of the difficult with problems like these is getting an accurate drawing from the description. That's something we can definitely practice as a group.

Assuming you arrived at the following:

You can deal with the easier part of the problem which is all about triangle area ratios. First lets remove the unneeded lines and split the parallelogram in half.

  •  First note ADF is half of the parallelogram and has an area of 7.
  • Then the area ratio of ADF : ACD is 6:33 
  • You then repeat this process: the area ratio of ACD : ABC is 7:26

Problem 13 

Suppose x and y are nonzero real numbers simultaneously satisfying the equations \( x + \frac{2018}{y} = 1000 \) and \( \frac{9}{x} + y = 1. \) Find the maximum possible value of x + 1000y.

My first instinct in these problems is to always remove the fractions to get:

$xy + 2018 = 1000x$ and $9 + xy = y$ Then just on inspection we have the 2 parts of the expression we want to simplify   \( x + 1000y = 2xy + 2027 \)

And we also have an way to isolate either x and y, I picked y, to plug them back in and get

$y = \frac{9}{1-x}$ =>  $x + 1000 \cdot \frac{9}{1-x} = 2x \cdot \frac {9}{1-x} + 2027$

That cleans up quickly to: \( 1000x^2 - 3009x + 2018 = 0\) and while the numbers are high the factorization is still not too hard (1000x - 1019)(x - 2).  So despite the phrasing which suggests an optimization problem there are only 2 solutions  for (x,y) and you just have to plug them in and compare to get the final result.

Problem 15

There are integers \(a_1, a_2, a_3, . . . , a_{240}\) such that \( x(x + 1)(x + 2)(x + 3)· · ·(x + 239) = \sum_{n=1}^{240} a_n x^n \). Find the number of integers k with 1 ≤ k ≤ 240 such that \(a_k\) is a multiple of 3.

At first glance this is a slog of a counting problem. Multiplying all 240 binomials together is impractical without a computer program.  So the first step I took was to look for patterns by doing the first few terms,

Ignoring x which doesn't change the coefficients:

$$(x+1)(x+2) = x^2 + 3x + 2$$
$$(x+1)(x+2)(x+3) = x^3 + 6x^2 + 11x + 6$$

I noticed a few things from this:

  • all the x+3n terms always added 1 more multiple of 3 coefficient than the last term. This makes sense the x term just shifts the previous result and then we add a multiple of 3 to it which doesn't change the value modulus 3 and finally we get one new multiple of 3 constant term at the end.
  • So grouping the terms looked interesting.  Note: there are 79 terms with  a multiple of 3 in them. So that's 79 coefficients with a 3 at the end as a minimum. 
  • Next: modular arithmetic seems useful here. We can simplify everything to mod 3 and not change the result so now we have x(x+1)(x+2)(x+3)(x+1)(x+2)(x+3) ....   I chose 3 rather than zero deliberately so we didn't lose any terms.
I also started simplifying the calculations letting x be an implicit place value system  and never carrying so

for example squaring the first term (x+1)^2  becomes (1 1)^2  = 1 2 1 

Somewhere around this point I also converted to balanced ternary and used  \( ( x+3)^{79} \cdot (x+1)^{80} \cdot (x-1)^{80} \)

That's convenient because we can easily expand the two final terms. At this point you're left with multiplying two highly symmetrical 81 digit terms.  (1 1 1 1 1 .... 1 1)(1 -1 1  -1 1 -1 ... . -1 1) 

As far as I  can see there's no escaping working out this product at least to a few terms before the pattern is visible but its a fairly simple process and you quickly see the result looks like (1 0 1 0 1 ... 0 1) with 161 terms and therefore 80 more zeros which are really coefficients that are multiples of three. That gives the baseline number of coefficients. So we add the two parts together to get 80 + 79 = 159. 


I went with another UWaterloo problem set this week: with a fairly approachable  number theory/lcm type problem.

Tuesday, April 10, 2018

Tricky Geometry Problem

In this walkthrough we start with one that didn't look too hard at first:

Goal: Find angle \(\angle \alpha \)  [Ignacio Larrosa]

It quickly became apparent though that was going to be tricky.

Avenues of investigation
1. Angle chase. The isosceles trinagle ADM is the starting point .At this point things were going well. All the angles can be described in terms of alpha and many are similar.

2.  Somewhere at around this point I decided to model in Geogebra to see what the goal really was. 
This gave the target 18 degrees which was helpful. It also showed that ABC was a right triangle. So one path of investigation was to look into whether we could find expressions for the side lengths and show they satisfied the Pythagorean Equation. But it was hard to get them in a common variable.

3. I started looking for similar triangles - BM  reflected across BH creates another isosceleses for instance in an effort to build up side lengths but couldn't derive the 3rd side without another variable.

4. I also added a few other auxiliary lines to see if they aided including splitting DAM in half, the perpendicular bisector of AC. I extended AB to create a larger outer isosceles triangle.

5. Played with Stewarts Theorem and various similar triangles trying to derive BM's length and looked for ways to show BM = AM.  That would prove B was on the circle curmscribing AC.

6. I then became interested in subdividing BAM and the cyclic quads there. This was the breakthrough  I found enough similar triangles in this division to create two expressions for a subsegment.  Even this idea took a bit of playing before it shook out.

Here's what I came up with in the end:

1. Angle chase and a lot of similar triangles and cyclic quadrilaterals fall out.
2. I subdivided angle \(\angle DBA\)  on the right so I had more combinations of angle alpha to play with. It also seemed useful to add the perpendicular bisector of AB.

Note: green angles are \(\angle \alpha\) and can be combined to make \(2\alpha\) or \(3\alpha\)  red angles are \(90 - 3\alpha\).  and the blue angle is \(2 \alpha\)

In particular: \( \triangle BIJ  \simeq   \triangle EHI \) forming a cyclic quad and therefore angle \( \angle EHB  \) is a right angle.

1.  So  let BC = AC = BG = m and   let  CJ =  a and BJ = m - a.  Then since \(\triangle EHI \) is some scaled version of \(\triangle IBJ\) lets assign k equal to the scale factor. Based on that EH = \( k \cdot BJ =  k\cdot (m -a)\)  Likewise EI is k scaled version of BI.

2.  Next note \( \triangle BEI  \simeq   \triangle ACE \) .  Since EI is a k scaled version of BI, CE is a k scaled version of AC or \(k \cdot m\)  and CH = CE - EH = \(k \cdot a \)

3. Also \( \triangle BIJ  \simeq   \triangle BCH  \simeq \triangle CEJ \) all  \(\alpha\) right triangles  and from this we get

\( \frac{CH}{BC} = \frac{CJ}{CE}\) or \(\frac{k \cdot a}{m} = \frac{a}{k \cdot m} \)

But after simplification this implies k  = 1   and \( \triangle BIJ \) is not just similar but congruent to
\( \triangle EHI \)

4. So EI = BI and \(\angle IBE = \angle BEI \)   This means \(2 \alpha = 90 - 3 \alpha \) and \(\alpha = 18 ^{\circ} \)

Wednesday, April 4, 2018

4/3 Spring is Sprung

The last few weeks I've been concentrating on arranging the topics up to the last MOEMS Olympiad. This week I looked forward enough to realize that first Spring Break was coming up and that immediately after it was our only window to do the Purple Comet Math Meet.  So I tabled my original idea for a new game in favor of doing a practice problem day.  To start things off, I had one new student join us so we went through introductions again. As usual I had everyone say their name, math class and favorite activity from the last few months (except for the new boy who I asked to say why he joined)  There were a couple of trends in what was mentioned. Pi day was apparently quite memorable and a group favorite as well was the Math Counts competition itself. But my favorite comment from someone a bit unexpected was to the effect "I really like all the problems, they're not like math class at all." 

Before we started I had the kids go over the problem of the week (from @solvemymaths) on the board:

I was happy to get both compute the inner area and compute the outer area demonstrated as approaches by two different students. (As I type this write-up I just notice the slanted 90 degree angle and wish we had talked about it.)

Since Purple Comet is a team event that you run in your own room I asked whether the kids wanted me to randomize the groupings or if they wanted to form them on their own.  There was a  very strong strong preference to choose teammates.   As no one was left out I let them do so although every time this happens I wonder a bit about random groupings and if I should occasionally impose them on the kids. So far my feeling is that as a voluntary club its ok to honor their wishes and that regular class will cover some of this ground. 

Once everyone was divided I used a structure that has worked well in the past.  I wrote the problem numbers up on the whiteboard and had the teams record their answers there. If they noticed a difference, they then went and discussed their solutions between teams to come to a consensus (and I also could focus on the groups at that point too.)

Even though its not really competitive this is just enough structure to hook a lot of the kids in and keep them going. I also brought some slant puzzles to hand out if I saw kids "burning out."  but I really only used them with 2 kids at the very end. Engagement overall maintained itself.

I had the kids go over last year's contest: 2017 MS Contest Based on yesterday it looks like everyone should be able to do at least 6-10 problems on the real event which for me means the levelling is pretty good this time.

During this whole process I then circulated and helped out with individual groups.  Interestingly, the one problem I ended up focusing on the most with all the groups the was the tower of 7's.

Find the remainder when \( 7^{7^7} \) is divided by 1000. 

There's not a lot of number theory or modular arithmetic exposure in school so this isn't so surprising.
I emphasized a few ideas:

  1. Find the pattern/cycle in the last 3 digits of the powers of 7 i.e. make a table and see what happens.
  2. You don't need all the digits to keep calculating only the last 3 (and why?)
  3. Once you know the cycle length you just need to find where you are in the cycle i.e. what's the remainder of 7^7 divided by 20.
  4. That can also be done by the just looking at the remainder (mod 20) after each multiplication of 7 and there is a simple pattern there as well.
I'm almost completely decided we will do something with modular arithmetic before the end of the year based on this experience. 

Monday, April 2, 2018

In praise of the Rational Roots Theorem

First some personal historical background. In my school district, you could do Algebra in middle school but unlike a standard class it only covered linear equations.  When I was in High School after doing  Geometry in 9th grade, I entered a 3 year accelerated math sequence that terminated with AP Calculus BC.  For the first year we did a semester going over quadratics leading up to the derivation of the quadratic formula and a semester of trigonometry.   So for all intents and purposes, I didn't learn anything from the standard Algebra II curriculum. Interestingly, this didn't have any particular consequences and as time went by I learned some of the topics when necessary and required for something else. I remember thinking in College, "I wish I had covered more Linear Algebra/Matrices" but never really "What's Descartes' rule of signs?"

However, over the last few years my affection for two particular tools from there has grown quite a bit: The Rational Roots Theorem and Polynomial division.  First, these are often under attack and dropped (just as in my own experience).  Its not unusual to see people wonder online: what are the real world applications of these or will they ever be used again?  In High School, I might have said you can always graph and use approximation techniques like Newton's Method when these come up. More significantly,  the existence of Wolfram Alpha has made generalized solutions to cubic and quartic equations easily accessible (if not derivable) From my perspective, they are two basic polynomial analysis techniques that offer a gateway to understand higher degree polynomials.  That understanding is valuable in itself but in addition they offer a fairly general technique for a lot of algebraic puzzles that I try out and I find it extremely satisfying to be able to analyze these with  just pencil and paper.

Example 1:

$$x^2 - 13 =\sqrt{x + 13}$$

This looks not to hard at first until you square both sides to get rid of the radical and realize its a quartic in disguise:

\(x^4 - 26x^2 + 169 = x + 13\) => \(x^4 - 26x^2 -x + 156 = 0\)

A common strategy at this point is to look for clever factorizations.  But its often really hard to see where to start. In fact, I find these are often easier to derive backwards after you know the roots anyway.

So let's start with a quick graph of the  functions. This could be done by hand but I'll use geogebra here. The left hand side is a parabola with vertex at (0,-13) while the the right hand side is half of the rotated 90 degree version of the same parabola with a vertex at (-13,0).  If you're looking for factorization this symmetry is something that provides an avenue of attack. But for our purposes it also shows us there are only 2 real roots in the quartic and approximately where they lie.

Here's where the Rational Roots test comes in. Since 156  = \(2^2\cdot 3  \cdot 13\) It says that if there is rational root its going to be either: \( \pm 1, \pm 2, \pm 3, \pm 4\, \pm6, \pm 12, \pm 13, \pm26,  \pm 39, \pm 52, \pm 78\) or \( \pm156 \)  That's a bit daunting but looking at the graph or the behavior of the functions indicate we really only need to test smaller values and 4 is probably the most promising.

I just plugged that back into the original problem rather than doing the quartic and indeed -4 works out. (This is a bit of a cheat since not all the quartic solutions are also solutions to the original problem due to sign issue with the radical but if it does work then you're golden.)

At this point we know know x+4 is a factor of the original quartic and we can divide it out to get a simpler cubic equation.

Apply polynomial division \(\frac{x^4 - 26x^2 - x + 156 }{x+4} = x^3 -4x^2 - 10x + 39\)  to get the remaining cubic part of the equation.

Now once again we can apply the rational roots test but on the much smaller set {1,3,13,39}.  Its clear from the graph that none of these are going to be a solution to the original problem and that again the smaller ones are more likely. So starting at 1, I find that 3 works out  (27 - 36  - 30 + 39 = 0).   That mean x -3 is another factor.  Interestingly you can see why it doesn't work:  9 - 13 = -4 while the square root of 13  + 3 = 4.  So the inverse sign changes have interfered (but if the bottom of the sideways parabola were present that would be an intersection point).

Once again apply polynomial division  \( \frac{x^3 -4x^2 - 10x + 39}{x - 3} = x^2  - x - 13 \)   Having factored the quartic down to an approachable quadratic we can now  apply the quadratic formula to find two more solutions:  \( \frac{1 \pm \sqrt{53}}{2} \).   Either by testing or looking at the graph we can see \( \frac{1  + \sqrt{53}}{2} \) is the second solution while its converse again lies on the intersection of missing bottom half of the sideways parabola.

Extension for another time:  We have 3 and 4 wouldn't it be nice if 5 also showed up (and this is tantalizingly close to the generator function for pythagorean triples in the complex plane)?  Is there a general form to the intersections of this type i.e. a parabola and its rotated counterpart?

Example 2:

Find the integer solutions to: \(x^3y^3 - 4xy^3 + y^2 + x^2 - 2y - 3 = 0\)

This again looks fairly complex and of degree 6 on first glance. But lets try experimenting with values of x and see what falls out:  [I'm going to only consider the x >= 0 for simplicity here but somewhat similar logic applies for the negatives.]

if x = 0 this simplifies to:

$$y^2 - 2y - 3 = 0$$ which has 2 integer roots.

if x  = 1 this simplifies to:
$$-3y^3 + y^2 -2y - 2 = 0$$
We can applies the rational root test and check the possible integers \(\pm1, \pm2\) with no hits.

if x = 2 this simplifies to:
$$y^2 -2y + 1 = 0$$ which has one integer solution.

Note the constant term flipped from negative to positive at this point and now something interesting happens

if x = 3 this simplifies to
$$15y^3 + y^2 -2y + 6 = 0$$
The rational roots test now is only going to give positive candidates and the higher degree terms start to dominate making it impossible for this to reach 0 with an integer. I.e. \(15y^3 + y^2  > 2y\) for all integers > 1.

Continuing on the same trend  if x = 4 this simplifies to
$$48y^3 + y^2 -2y + 13 = 0$$  For the same reason this is even worse  \(48y^3 + y^2  > 2y\) for all integers > 1.

So we can infer with this logic that no integer solutions exist above x = 2.

Example 3:

$$x^2 - xy + y^2 = 13$$ $$x -xy + y = -5$$

Note: this is the symmetric intersection of a tilted ellipse and hyperbola.  There's a clever substitution that can simplify this which I'll mention at the end but in the general case this system is actually a quartic in disguise.

If directly attacking the problem, the starting point is the second equation with the simpler degree terms that allow us to easily isolate x or y.

\( x + 5 = xy - y = y (x -1)\) or after confirming x - 1 => x = 1 is not a solution and can be safely divided  \( y = \frac{x + 5}{x -1}\)

We can substitute that back into the first equation to get:

$$x^2 + x \cdot \frac{x+5}{x-1} + (\frac{x+5}{x-1})^2 = 13$$

Now multiply everything by \((x-1)^2\) to arrive at:

$$x^4 - 2x^3 + x^2 - x^3 - 4x^2 + 5x + x^2+ 10x + 25 = 13x^2 - 26x + 13$$ which cleans up to the following quartic:

$$x^4  -3x^3 - 15x^2  + 41x + 12 = 0$$
Again using the rational roots theorem you only have to test 1,2,3,4,6, and 12. I also check from the bottom up since its simpler and you don't need to recheck again if you find a factor and divide to a simpler polynomial. Going up from one, first 3 and then 4 test successfully as roots leading to the following factorization once you do the polynomial division:

$$(x-3)(x-4)(x^2 + 4x + 1) = 0$$ and from here the four (symmetric) solutions fall out.

As promised the clever factorization in problem like this where there are only xy and x + y terms is to consolidate and substitute:

First rearrange the original equations:  \((x+y)^2 - 3xy = 13\) and \((x + y) + 5 = xy\)

Substitute for xy to arrive at \( (x+y)^2 -3(x+y) -28 = 0\)

Then substitute again w = x + y and solve the quadratic  \(w^2 -3w - 28 = 0\) Once done replug the solutions x + y = 7 and x + y = -4 back into the original equations to solve a second set of quadratics. Note: this is actually probably just as much work as directly attacking the quartic since in effect you have to handle 3 quadratics.

Further Digression:

Conceptually I actually prefer the following approach.  From either the graph or the symmetrical nature of both equations it looks interesting to rotate them 45 degrees back to non-tilted form. This is desirable because it will remove xy terms.

So applying the standard rotation equations:  $x' = x cos(\theta) + y sin(\theta)$ $y' = x sin(\theta) - y cos(\theta)$ to the two equations for \(\theta = -\frac{\pi}{8}\)  we get:

\(x^2 - xy + y^2 = 13\) becomes \((\frac{\sqrt{2}}{2}x - \frac{\sqrt{2}}{2}y)^2 - (\frac{\sqrt{2}}{2}x - \frac{\sqrt{2}}{2}y) (\frac{\sqrt{2}}{2}x + \frac{\sqrt{2}}{2}y) + (\frac{\sqrt{2}}{2}x + \frac{\sqrt{2}}{2}y)^2 = 13 \)

That simplifies to \((x')^2 + 3(y')^2 = 26 \)

Likewise \(x - xy + y = -5\) becomes \( (\frac{\sqrt{2}}{2}x - \frac{\sqrt{2}}{2}y) - (\frac{\sqrt{2}}{2}x - \frac{\sqrt{2}}{2}y) (\frac{\sqrt{2}}{2}x + \frac{\sqrt{2}}{2}y)  + (\frac{\sqrt{2}}{2}x + \frac{\sqrt{2}}{2}y) = -5 \)

Once again this simplifies nicely to \( (x' - \sqrt{2})^2 - (y')^2 = 12\)  and these two equations are solved in a fairly standard fashion via elimination i.e.

\(   3(x' - \sqrt{2})^2 - 3(y') ^2 = 36 \)
\(+    (x')^2 + 3(y')^2  = 26 \)
\(    4(x')^2 - 6\sqrt{2}x'  + 6 = 62 \)

Solving with quadratic equation and \(x' = \frac{7\sqrt{2}}{2},  y' = \pm \frac{\sqrt{2}}{2} \) or \( x' = -2\sqrt{2}, y' = \pm \sqrt{6} \)

If you look back at the previous solutions at first this looks odd but remember these have to be tilted back 45 degrees to give the proper solutions and sure enough for instance \( (\frac{\sqrt{2}}{2} \cdot \frac{7\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \cdot \frac{7\sqrt{2}}{2} -  \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}) = (4,3) \)

Friday, March 30, 2018

3/27 Olympiad #5

As is our routine, I started by going over the problem of the week:

Find six distinct natural numbers A,B,C,D,E,F such that

A + B + C = D + E + F


Once again, I had a girl write a python program to find the solution. I love all the computational computing that is occurring. I need a better way to harness this energy. This time I had her talk about the structure of the loops she used to search the problem space.  In this case, it was a straightforward brute force attack loop over all 6 variables up to a limit and just check for each permutation if the two conditions were met.

This is relatively slow and it produced lots of duplicate solutions as she pointed out that had to be manually disambiguated. I didn't go into it because not enough of the kids have a programming background but there are a few easy improvements that can be made on this approach:

limit = 100

for total in range (6,limit):
    results = dict()
    solns = 0
    for a in range (1,total - 3):
        for b in range (a+1, total - (3 + a)):
            c = total - (a+b)
            if c < b:
            sum = a*a + b*b + c*c
            if sum in results:
                solns += 1
                print "Found %d %d %d and %s total:%d" %(a, b, c, results[sum], total)

            results[sum] = (a, b, c)

    ratio = float(solns) / total
    print "%d solutions for %d ratio=%f" % (solns, total, ratio)

Looping over the sum and then over the 3 digits in increasing order eliminates duplicates and cuts the number of comparisons down significantly.  Note: the key observation is that you only have 2 degrees of a freedom once you've picked a sum for a + b + c.

After this point, we did the final MOEMS olympiad for the year. I delayed this round to fit better with the other activities I wanted to do.  So technically we only had this week to finish and submit the scores.  Two things stood out at me. There was a fraction question that dovetailed with the 2 weeks we've worked on Farey Sequences and Wilf-Calkin trees. Also yet again there was another combinatorics problem that most kids enumerated over rather than calculating a true combination.  Overall, I think the kids scored the highest average of all the rounds. This afforded me the opportunity to have to cold call a few kids that usually are more reticent to demo on the board.  One highlight for me was  a girl proudly getting all the problems right and telling me that she thought this one was easy.  I know she meant relative to the other ones for her but I still had to make a comment to avoid language like "this is easy."  Nevertheless that was  a huge victory. 

I chose a page from the "This is not a math book" for the early finishers with a fun rectangle dividing project. (I brought my box of crayons in for this.)

This was popular but didn't occupy as much time as I expected and I ended up giving out the P.O.T.W early to some kids as well as breaking out my game of 24 cards.


This one comes from Ed Southall and is a fraction  talk activity:

Wednesday, March 21, 2018

3/20 Visible Math

This week started with a walk through of the MathCounts problem that I gave out last week to do at home.

Six standard six-sided dice are rolled, and the sum S is calculated. What is the probability that S × (42 – S ) < 297? Express your answer as a common fraction.

This was the last question in the sprint round at Chapters.  As I remember from the stats almost no one at the entire contest finished it correctly making it the hardest of the set.  I decided this would make for a good communal walk through because so many of the kids had seen it once and it hits a couple of different themes.   However, that's also the weakness of this problem. Conceptually its a bizarre hybrid  of a counting problem and a quadratic inequality neither of which naturally goes with each other.   I actually mentioned this to the kids. The phrase "franken-problem" might have been used.

At any rate, I started with the basics and asked some background questions:

  • What is the range of values for the sum of the dice throws?
  • How many total combinations are there for 6 dice throws in a row?  Why?
  • What is the most common sum / what would a probability graph look like?
This part was very approachable and the kids easily supplied various answers. So it was time for the quadratic inequality.  First I asked how many kids knew how to solve this algebraically? (Some of the room have not covered this at all)  It turns out even those kids with Algebra actually used guess and check anyway. There are only 31 values after all and its not too hard to just plug them in and see what happens.  The risk here is missing there is a range at both ends of the curve which I mentioned.

I had one volunteer who brought the equation into almost standard form but no volunteers to finish the process. So I demoed the formal method myself.
  • Factor  to:  (S-33)(S-9) > 0
  • Do a parity check: both factors are positive in which case S > 33 or both factors are negative in which case S < 9.
  • Notice the symmetry.
This felt new to the room and the work with signs of the inequality also exposed some conceptual weakness. So something to look for more problems to do in another context.

From here the problem becomes more standard and I had the kids do the case work on numbers of combinations for the 2 ranges.  We've been doing small amounts but could also use more combinatorics exposure.

That covered, I was ready for the fun part of today.  I've been looking at George Hart's site and was fascinated by some of the constructions. So I chose the sample one: to try out. 
Over the weekend I tested the templates and built my own ball:

It was a bit tricky, my ball almost fell apart at one time and I misplaced a few triangles leading to a dead end all of which gave me some ideas for how to guide when the kids tried it out.  Its really important to stress being precise when cutting the slots and also to work together when building the ball out to hold it together.

Beforehand I pre-printed the templates at a copy shop on 110 lb card stock paper. I also bought some thicker colored card stock which couldn't go through a copy machine and required tracing. I then mostly followed the lesson suggested on George Hart's site. We worked through discovering combinations of 3, 4 and five triangles first before really working as group. It took the kids the entire rest of the hour to build the balls once in white and then again in a multicolored version. 

This last one above was the most hard fought version. This group was the least focused and sloppiest cutters. So there were a few weakened triangles in their set. I kept coming over for a bit and helping them move forward with advice for kids to help hold the structure in place etc. But then in between when I went to work with others it tended to collapse.  Finally, I decided I really wanted everyone to achieve success and I should stay in place until they finished. I had them substitute in some borrowed extra triangles from the other groups and basically guided them through the tricky middle stage when the ball is most unstable. They finished right at the end and there was a literal cheer from the group. (I was extremely relieved)

The other groups actually made it through the multi-colored version where I had them try to create a symmetry in their use of color:

I was hoping to have enough time to discuss the extension questions about the combinatoric aspects of the colored balls but we ran the clock down.  As usual for me, I worried about the exact opposite case and had printed out the next template for early finishers which no one needed to use.   I'm currently testing this at home.  (Someone has to use the card stock.) Based on that experience the second ball is quite a bit more difficult to assemble and I'd budget much more time for it / prepare for some dexterity challenges.  That said, overall, I highly recommend this project. It was definitely a crowd pleaser!

(Its a bit like the 2nd death star right now)

This one comes from Matt Enlow and is an interesting number theory experiment.

Wednesday, March 14, 2018

3/13 Pi Day - 1

This is my fourth experience with Pi Day or "Pi Day - 1" as I called it since we meet on Tuesdays.



In a nutshell, because there's pie to eat, the kids always have fun.  But I was reminded of another perspective today from @evelyn_lamb

"Pi Day bothers me not just because it celebrates the the ratio of a circle’s circumference to its diameter, or the number 3.14159 … It’s also about the misplaced focus. What do we see on Pi Day? Circles, the Greek letter π, and digits. Oh, the digits! Scads of them! The digits of π are endemic in math gear in general, but of course they make a special showing on Pi Day. You can buy everything from T-shirts and dresses to laptop cases and watches emblazoned with the digits of π."

Image result for larry shaw picture exploratorium

I'm pretty much in total agreement with above. I've gently ranted in the past about pi digit memorization contests and other such trivialities.  But as her article continues, there was a man behind the holiday, Larry Shaw the recently deceased director of the San Francisco Exploratorium. I think his vision was more than just eating pie but it was also an incredibly whimsical gesture which is why I believe its had as much cultural resonance.

So I take the day partly in that spirit of whimsy and also with the mission to always ground it in circle geometry in some way and as said at the start, the kids always have fun celebrating.  Mathematics doesn't have enough moments like this especially in school.

This year I decided to go back to the  basics. I had initially toyed with talking about the unit circle and the derivation of radians versus degrees but on reflection I found  so much material that I couldn't fit that in.  Instead, I started with a survey of student definitions of pi (while they were eating).  This was surprisingly solid. The phrase "ratio of circumference to diameter" came up almost immediately. I then took a poll of how many kids had already done activities in class where they measured circular objects of some sort and divided them  by their measured diameters to find pi approximations.  Again, almost everyone had done so often several years ago in Elementary School.

So with everyone convinced already pi existed and it had a value it was time for some deeper questions. The first one I posed was "Is measuring a single object a good way to prove pi's existence?"  We chatted a bit about accuracy and sample sizes as well as whether from a mathematical perspective we can ever prove something from samples. My favorite version of this is
"What if only ordinary people sized circles have a ratio around pi and if we could measure microscopic or macroscopic versions we'd find something different?"

One of the kids then suggested approximating the circumference of a circle with polygons so we then did that on the board for the hexagon version.  I cold called in this case which I usually don't do to get a student to sum up the perimeter of the hexagon arriving at pi is approximately 3.

From there we took a quick digression to also do the area of a circle visual proof where you cut the circle up and form a rough rectangle that is pi*r by r in size. Again I had the kids fill in and compute the area.

Finally I noted that we don't actually compute pi to a billion digits using geometry and asked if anyone knew of other ways to get it.  This was a new idea for the room and a good setup for the 2 videos I chose for the day.

The first was this amusing (there were a lot of genuine laughs while watching) video of Matt Parker computing pi by hand using the alternating series 1 - 1/3 + 1/5  - 1/7 ....

But of course this doesn't really explain why this works only that it appears to do so. So I also picked the very ambitious following one by 3blue1brown:

Its about as approachable as its going to get with this amount of background knowledge but still a stretch. I stopped several times to ask questions about some of the background concepts. There are several potential stumbling blocks here:

  1. law of inverse squares
  2. Inverse pythagorean theorem
  3. The general abstraction model used
  4. The number line can be thought of as a curve.

The last one was the one  I chose to focus on the most and I framed it as a thought experiment "What if the number line isn't really a line at all but a curve, we're just at a small portion of it and just like with a curve if you magnify enough it appears to be straight."  My hope is that if nothing else stuck that idea was interesting and thought provoking (hello Calculus in the future)   My informal survey is that most kids found it interesting but I may have had one where this pushed too far.  So I am planning to do a little preamble next week "Its ok to give me feedback if you found anything too confusing and I also sometimes want you to focus on the big ideas in moments like this even if the  details aren't accessible yet"

I gave out the last problem from MathCounts this year now that it was released:  

Its actually a fairly awkward merge of quadratic inequalities and dice counting problem but I wanted to provide a capstone to the kids experience there and dig into how to solve it.