Tuesday, December 20, 2016

Order of operations (a mini-rant)

Order of operations is a major pet peeve of mine for two reasons:
(1) Some people love to use it in "gotcha" challenges to make other people feel bad about their math abilities. Here are some examples: facebook meme, a similar one, and this:

(2) For some students, it stands as a clear example of the idea that math is a set of arbitrary rules they have to accept and/or memorize (It isn't!)

I think there is a very different way of approaching this issue which is much more mathematically rich and fun.

The first and most obvious is to treat these memes as games and see how many answers you can justify by making the order of calculation explicit (use parentheses). Implicitly, this is the idea behind games like 24 or the traditional New Year's challenge (use the digits of the new year to make all values from 1 to 100).

2. Talk about history
Some launching questions: Where did the order of operations come from? Is it universally agreed?
This article from Tara Haelle does a good job of talking about this perspective and gives some further references.

3. Ask students for their own thinking
What do they think the order of operations should be? Why?
A tantalizing question: should there be an agreed (implicit) order of operations at all?

4. Talk about redundancy and error flagging (and error correcting) codes
Jordan Ellenberg's How Not to Be Wrong has a great discussion about redundancy in language and related code concepts. One key point is that we always face a trade-off between brevity and transmission errors. In other words, we can write short messages where every character carries critical, independent information, or we can use a system in which our messages are longer, but carry duplication and internal references that make our meaning more robustly clear. (compare the previous two sentences!)

Relying on a convention, like the order of operations, means that we can use fewer symbols to convey a mathematical expression. The great danger comes if the author and the reader don't share the same conventions! A less obvious danger is if a symbol gets garbled in the transmission, it may be hard to identify the error or even see that there was an error.

Monday, December 12, 2016

Some sort of number talks with J3

Based on conversations about the dots pictures from Math4Love:

Day 3
I notice

1. there's a number 3, but the number of dots isn't the same (it isn't 3)
2. five over here (pointing to dots) and zero on the down part (the bottom half of the 10 frame)
3. J0: I see some letters...
4. I even noticed that. I noticed there's this plus (points to dash - )
5. J0: I noticed this square
6. I noticed it was a line (bottom row of the 10 frame)
7. I noticed these triangles (the white space in 10 frame sections that have dots)
8. I noticed these are 5 and an extra one (on second page of day 3)
9. I noticed that there are four left (empty cells on second page)
10. J0: you saw 5+1, I see 2 + 4
11. Those two are together. The other ones are lonely.

I wonder

1. Why didn't they make 10 dots?
2. Why did they only cover the middle of the square (points to a dot in the upper left square of teh 10 frame)?
3. I wonder, how do numbers talk? (after I read the title of the slide to her)
4. I wonder, why do they only put 1 on the bottom row?
5. I wonder, can we arrange them so none are lonely
Day 4
I notice
1. Five on the top and five on the bottom
2. Ten
3. five and four, nine
Day 5
I notice
1. This doesn't have a box to go in (a 10 frame)
2. It has a dot in the middle
3. we can count them 2, 2, 2 (pairing them up)
4. if we take 2 away, there will be four
5. the sides are the same (it has a line of symmetry in the middle)
6. it looks like an animals footprint
7. the top four make a diamond
8. If we turn our body to the side, the top four make a rectangle
9. taking out the two in the middle, we have a square
10. it has 8 dots.
11. the number of dots doesn't match the day number
I wonder
1. is it a real footprint?
2. I wonder, if we take the bottom five, it would be 3?

Thursday, December 8, 2016

Solving contest problems (challenge from Mike Lawler)

These are my notes working through problems posted by Mike Lawler on his blog. You'll have to go there to see the problem statements.

The intended value of this write-up is to show examples of the actual problem solving thought process someone has followed, not just a polished solution.

Problem 19
Since the circle has area 156 π the radius squared is 156, which is 4 * 3* 13. That seems like a strange number, I'm curious to see where it will make calculations come out nicely, later in the problem.

We're told OA has length 4 sqrt(3). Squaring that only gives 48, so A is well within the circle. That means triangle ABC has vertex pointing toward O along the perpendicular bisector of side BC.

This lets me set up a picture of a right triangle with legs length $x$ (half the side of the equilateral triangle), $x$ sqrt(3) + 4 sqrt(3) (the altitude of the equilateral triangle plus OA) and hypotenuse r (2 sqrt(3*13)).

Applying the pythagorean theorem and simplifying along the way:

$$x^2 + 3(x+4)^2 = 156$$
$$4x^2 + 24x + 48 - 156 = 0$$
$$x^2 + 6x - 27 = 0$$

Visually factoring gets me $x$ is either -9 or 3.  3 is much more reasonable for half the length of the side of a triangle, so my answer is 6.

Note: If this weren't a contest problem, I might try to think about the -9 root a little more carefully.

Equiangular hexagon
I had already solved this problem from a recent tweet of Mikes, so I can't fully recreate my thought process.  Here were some of the highlights:
• wonder why 70%
• draw a picture: fail to notice that triangle ACE is equilateral
• Split the hexagon into two trapezoids by line CF
• Calculate the area of those two trapezoids
• Calculate the length from C the intersection with AE and segment CF.
• Calculate the area of ACE based on the two subtriangles split by CF.
• Obtain the quadratic equation for r based on the formulae for the two areas and the given 70% parameter.
School Competition
Let's call Andrea's rank $m$ for median. Since she is the unique median, there must be $2m - 1$ total contestants. We also know:
• $m \leq 36$ because Andrea scored higher than Beth who was ranked 37th
• $64 \leq 2m -1$ because Carla ranked 64th, so there were at least that many competitors
• $3 \mid (2m - 1)$ since every school sent three competitors
The first two inequalities tell us that $33 \leq m \leq 36$. Because $2m -1$ is a multiple of three, $m$ can't be a multiple of 3, so it has to be 34 or 35. Calculating mod 3, we can quickly check both and see that $m$ has to be 35.

That means the competition had 69 competitors from 23 schools.

Using multiple choice
After putting together these notes, I saw a commenter on Mike's blog use "solution by multiple choice." When I was doing timed tests/contests, I would make use of the options as part of my strategy. However, I don't do that now, since I'm much more interested in exploring the mathematics of the problems than getting the final answer.

Monday, December 5, 2016

Leftorvers with 100 game

In Grades 3 and 4, we played a nice game that (I think) we got from Marilyn Burns. Looking for a reference after the fact, I see it explained in her book Lessons for Extending Division.

Basic play

• Start with a target number (we used 100) and collection of available divisors (we used integers 1 to 20)
• Players take turns choosing a divisor from the remaining available options. They divide the current target by that divisor and keep the remainder as their score for the turn. They also subtract the remainder from the target to create a new target for the next player.
• Each divisor gets crossed out when it is used, so it can only be used once.
• The game ends when the target is reduced to 0 or when all available divisors are exhausted.
• We played as a two player game.

Here's an example of a game play:
Player 1 chooses 17. 100 = 17 * 5 + 15, so player 1 scores 15 points, the target is reduced to 85, and 17 is no longer available as a divisor.

Player 2 chooses 20. 85 = 20 * 4 + 5, so player 2 scores 5 points, the target is reduced to 80, and 20 is no longer available as a divisor.

Player 1 chooses 14. 80 = 14 * 5 + 10, so player 1 scores 10 points, the target is reduced to 70, and 14 is no longer available as a divisor.

Player 2 chooses 18. 70 = 18 * 3 + 16, so player 2 scores 16 points, the target is reduced to 54, and 18 is no longer available as a divisor.

Player 1 chooses 19. 54 = 19 * 2 + 16, so player 1 scores 16 points, the target is reduced to 38, and 19 is no longer available as a divisor.

Player 2 chooses 13. 28 = 13 * 2 + 12, so player 2 scores 12 points, the target is reduced to 26, and 13 is no longer available as a divisor.

Player 1 chooses 15. 26 = 15 * 1 + 11, so player 1 scores 11 points, the target is reduced to 15, and 15 is no longer available as a divisor.

Player 2 chooses 16. 15 = 16 * 0 + 15, so player 2 scores 15 points, the target is reduced to 0 and the game ends.

Player one wins 52 to 48.

Our experience
We found this to be a fun, interesting, and engaging game. The practice with dividing and remainders was pretty obvious. In addition, it opened up some opportunities for strategic thinking, particularly at the end-stage of the game. I think there are also several good extension explorations.

Extensions
First, I created a simple pencilcode program for two players to play this game against each other. Here's a playable version (and here's the code).

Second, you'll notice that the first player in our sample game followed a "greedy strategy."  At each stage, that player chose the divisor that would give the most points on that turn. If you look closely, that isn't the best strategy at the end of the game.

So, a natural exploration is to find the best strategy for different starting targets. One specific point of about which we're curious: is it ever desirable to skip your turn (choosing 1 as the divisor is effectively a turn skip)?

Some other areas for investigation:
• must the game always end on 0 or can we run out of divisors?
• given a target and collection of starting divisors, what is the shortest (number of turns) game possible? What is the longest game (number of turns) that does end at 0?