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CAS Conference

Computer algebra systems (CAS) have the potential to revolutionize mathematics education at the middle and secondary level. Experience how CAS can be integrated into Pre-algebra, Algebra 1 & 2, Precalculus, Calculus, Statistics, and Geometry classes.

Consider attending (or maybe even proposing a session) for the 9th USACAS conference, July 18-19, 2015 at Hawken School in Cleveland, OH.

• Discover how secondary and middle school teachers are using CAS in their own classrooms.
• Get classroom tested ideas developed for CAS-enhanced classrooms.
• Learn what other countries are doing with CAS.
• Interact with prominent US and International CAS pioneers.

Information on registration, speakers, presentations, Saturday outing, and hotel can be found at http://usacas.org .

Consider proposing a session at http://bit.ly/USACAS9_speakers .

Star angles

Given a 5-pointed star like the one shown below, what is the sum of the angles in the outer “arms” of the star?

A variation on this problem showed up on a contest our students took last week.  I think there are multiple solution approaches to this, but the exploration and discovery are well worth it.  I offer a generalization at the end.

While not permitted on the contest, my first instinct was to construct the star using dynamic geometry software.  The image below is my result using the TI-Nspire; other softwares would work equally well.

After dragging around the star tips, watching the angles change while the sum didn’t, it was pretty clear that the sum was fixed at 180 degrees.  Nice, but why?

Here’s my no-technology solution from the day of the contest.

I saw the figure as an interior pentagon with exterior triangles off each edge.  Using this, I knew that ALL of the angles in the figure must add to $3\cdot 180$ (for the pentagon) plus $5\cdot 180\deg$ (for the 5 triangles), for a total $8\cdot 180$ degrees.  Then I noticed that there were 10 linear pairs of angles where the triangles met the pentagon, two at each pentagonal vertex.  Subtracting these from the initial sum would leave just the desired star point angles, but unfortunately would subtract the interior pentagonal angles twice, so the pentagon would need to be added back to re-balance.  That left the measure of the start point angles for any orientation to be:

$8\cdot 180$ (0riginal) $-10\cdot 180$ (10 linear pairs) $+ 3\cdot 180$ (rebalance pentagon)

For a total of $1\cdot 180 = 180$ degrees.

I’d love to hear other approaches.

EXTENDING:

There’s some lovely geometry and arithmetic overlaps in the construction of stars.  That’s perhaps another post for another time.  For now, I’ll extend this to stars with 5 or more points as stars with fewer points aren’t possible with non-degenerate internal polygons like above.

Using the same logic as above, imagine an n-pointed star comprised of an interior n-gon with its edges extended to form n exterior triangles.  My goal now is to find the sum of the n exterior star point angles.

As before, the n triangle angles sum to $n\cdot 180$ degrees and the interior n-gon’s angles sum to $(n-2)\cdot 180$ degrees for a total interior angle measure of $(2n-2)\cdot 180$ degrees.

There are $2n$ linear pairs at the junctures between the n-gon and its exterior triangles.  Subtracting their sum, $(2n\cdot 180)$ degrees, from total angle sum subtracts the n-gon’s angles twice, so adding the n-gon’s angles, $(n-2)\cdot 180$ degrees, back in once gives the desired sum.

$(2n-2)\cdot 180 - 2n\cdot 180 + (n-2)\cdot 180 = (n-4)\cdot 180$ degrees

So, for any n-pointed star with $n\ge 5$ as defined above, the sum of the angles in the star’s exterior arms is $(n-4)\cdot 180$ degrees.  When $n=5$, we get the star point angle sum of 180 degrees from earlier.

This formula is itself not so interesting, but that one can know this general result from manipulating nothing more than the angle properties of triangles, polygons, and linear pairs is nice and completely accessible to geometry students of many levels and ages.

Again, I’d love to hear (and will post/share) any other solution approaches.

Optimization in Four Colors

I suspect many (most?) geometry teachers know of the Four Color Theorem (FCT) which roughly states that any flat map, no matter how complex, containing only contiguous regions with finite perimeter can be colored with no more than four colors with the only restriction for coloring being that regions have different colors if they share any boundary beyond a set of finite points. While the FCT is not a particularly useful to cartographers, it has historical significance as the first significant mathematical proof to have been established with extensive use of computers.

THE PROBLEM:  In secondary geometry classes, the FCT is typically just a footnote or factoid, but it is pretty easy to understand for students of all levels.  This year I decided to make it more interesting as an optimization problem.  If each color you use has a different “cost” per area unit, can you color a given map as “cheaply” as possible?

[I considered a maximum cost map, too, and convinced myself that the maximum cost map is just a flip of all the colors, assuming the change in cost is the same between all colors.  With that thought, saving money seemed the more “realistic” goal, so I went with minimum cost.]

MOTIVATIONS:  Perhaps the BEST part of this project was that I was not–and still am not–convinced that we have found THE optimal solution.  I was reasonably certain that I could determine a very good mapping cost, but the sheer number of possibilities would require significant computer run time and coding abilities (just like the original FCT proof!) to ferret out the best answer–resources not available to those solving the problem (the computing problem is an issue my school is actively addressing).  I loved having a problem in math where determined students might best their teacher–and some did!!  I also liked that this project significantly motivated my students to use spreadsheets to track their data–a different math resource than most were accustomed to using.

IMPLEMENTATION:  Experimenting, I decided to offer different versions of the coloring challenge to my 4th-5th grade math club and all of my 8th grade math classes (prealgebra, algebra, & geometry).

Project 1:  Our 8th grade humanities course had an Africa unit earlier in the year, so I returned there by asking all of the students to color this map of Africa.

We provided this spreadsheet of country names and areas along with these coloring costs:  Purple = $2.00/mi^2, Yellow=$2.50/mi^2, Red = $3.00/mi^2, and Blue=$3.50/mi^2.  After some discussions on the first day, the “border” rule was revised to note that countries whose borders were only large lakes (Democratic Republic of the Congo & Tanzania plus Chad & Nigeria) could be considered “not touching” for this project.

Political incorrectness confession:  We noticed a day after we assigned the project that we had inadvertently left off the relatively new South Sudan. I decided to leave the two Sudans as a single country for this exercise (thus the inked in portion of the map).  Having compromised the previous day on the lake-bordered countries, my error accidentally made the largest and 3rd largest African countries border each other–a nice confounding problem, I thought, for forcing students to determine which would get a cheaper color.

Project 2:  I gave the relatively simpler map of the lower 48 US States to our 4th-5th grade math club with coloring restrictions Red=$1.00, Yellow=$1.25, Blue=$1.50, and Green=$1.75.

RESULTS:  For the submission, students (in ones or pairs) had to submit their colored map, a spreadsheet showing their computations, and 1-3 paragraphs explaining their general coloring strategies, and especially how they handled the inevitable situations where their coloring strategies self-conflicted.  In general, we could have done a better job preparing students for the written portion, but the two most commonly stated strategies were

1. (Low level) We colored the biggest countries the cheapest as far as we could, and then colored the next largest using the next cheapest color, etc.  If we ran into conflicts, we “worked it out”.

2. (Stronger) Noticed after trying the obvious strategy above that the countries colored the 2nd cheapest surrounding a “cheapest color” country often had more area than the “cheapest color” country.  By paying a little more for the largest country, they more than made up for the added expense by coloring a collection of countries that in total had more area.

3. A  few members of my math club addressed some specific strategies like the 11-state ring of US states (MO-IL-IN-OH-WV-VA-NC-GA-AL-MS-AR) surrounding Kentucky & Tennessee made it possible to use just two alternating colors over a large area.

Using our color schemes, excellent scores for the US map were very close to, but just over $1,000,000. The best Africa map scores we found were just under$17,000,000.  As I noted earlier, I’m not at all convinced that we have found the optimum values, but part of the fun of these projects was that anyone with some calm logic and determination could break through.  My second best coloring scheme was from a student who had been exposed to the least amount of math.  If you can beat these scores, please share.

VARIATIONS:  After playing with this for a while, I’m convinced that all optimal solutions depend on the gap you set between the color costs.  The more expensive the next color is, the more motivation you have to not change colors.  I haven’t tried it, but I think strategy #2 above could be exploited more often if paint color jumps are smaller on a large, complicated map.

I’m also convinced that the initial paint cost is irrelevant.  It will change the total cost of the project, but it would just scale all values up or down.

I didn’t play with different step values in paint cost, but I can see that potentially changing the game, especially if the cost jumps increase as you approach the 4th color.

Enjoy.

Integral follow-up

One HUGE reason to ‘blog is to get feedback on your ideas.  Thanks to David’s response to my last post on my attempts to integrate $\int x^2\sqrt{x^2+1} dx$, I got another lead on how to tackle the problem –and– a little (comforting) confirmation that the integral really was difficult.

The suggestion that integrals of odd powers of secant might cycle was all I needed.  I ignoring the hint that Wikipedia might hold a solution to find a way for myself. Here’s what I found.

To compute $I=\int sec^3 \theta d\theta$, let

One application of Integration by Parts, a trig identity, and a cycling integral gives

Similarly, to compute $I=\int sec^5 \theta d\theta$, let

Using the result of $\int sec^3\theta d\theta$, Integration by Parts once, a trig identity, and a cycling integral gives this solution.

From the first post, if $x=tan\theta$, then $\int x^2\sqrt{x^2+1} dx = \int tan^2\theta sec^3\theta d\theta$. A Pythagorean trig identity turns this into a linear combination of the previous two integrals.  The problem that initially stumped me has now been solved using using only circular trig and other integration techniques!

With potential domain restrictions, the initial substitution $x=tan\theta$ with a Pythagorean trig identity gives $sec\theta=\sqrt{x^2+1}$.  Therefore,

one of the original solution forms, found this time completely independent of hyperbolic trig.  Thanks for the suggestion, David!

Volumes of Revolution and Differentials

Here’s classic volume of revolution problem:

Determine the volume of the solid created when the region defined by $y=x^3$, the line $x=2$, and the x-axis is rotated about the x-axis.

Typical calculus classes present TWO ways to compute such volumes of revolution–orthogonal and parallel to the axis of rotation.  In this post, I argue that with substitutions involving differentials, you can actually get your choice of FOUR integrals to solve.  Hopefully one of them will be easy to solve.  Here’s how.

Slicing the volume ORTHOGONAL to the x-axis creates discs with radius $y$ and thickness $dx$ giving an original (orthogonal) volume integral $\int \pi y^2 dx$.

Method 1) I think the approach most familiar to most students is to substitute using $y=x^3$ to get

Method 2) Here’s my suggestion that I don’t see too many people use.  Consider substituting for the differential using $x=y^{1/3} \longrightarrow dx=\frac{1}{3} y^{-2/3} dy$ to get

Using the method of cylindrical shells, slicing the volume PARALLEL to the axis of rotation creates a series of annuli, each with radius $y$, thickness $dy$, and height $(2-x)$, giving an original (parallel) volume integral $\int 2\pi y \cdot (2-x) dy$.

Method 3) Substitute $x=y^{1/3}$ into this to get

Method 4) Or you could use the other differential, $y=x^3 \longrightarrow dy=3x^2 dx$, to get

CONCLUSION:

No matter what approach you take, it’s important for students to understand that there is no single approach that will will work well for all volumes.  Increasing the number of representations available for manipulation generally should increase your chances of finding one that’s easy (or at least easier) to solve.

In the end, I just think it’s pretty that all four of these integrals are exactly equivalent even though they really don’t look much alike.

Clever math

Here are what I think are three clever uses of math by students.

In my last week of classes at my former school in May, 2013, my entire Honors Precalculus class showed up wearing these shirts designed by one of my students, M.  The back listed all of the students and a lovely “We will miss you.”  As much as I liked the use of a polar function, I loved that M opted not for the simplest possible version of the equation ($r=2-2sin(\theta )$), but for a rotation–a perfect use of my transformations theme for the course.

Now for a throwback.  When I was a graduate student and TA at Syracuse from 1989-1990, one of my fellow grad students designed this shirt for all of the math and math ed students.  I don’t remember who designed it, but I’ve always loved this shirt.

Nspire Apps Update

Thanks to a great comment from Mr. Schirles on my post on the new Nspire app from earlier today, I’ve learned something new.

When you are using the math keyboard in the app, notice that several of the keys have overlines along their tops.  When you tap-and-hold those keys, you get access to several related keys.  For example,

• Holding the r-key gives $\theta$.
• Holding > gives the other three inequalities.
• Nicely, holding the sine button button reveals the related arcsine, cosecant, and arc-cosecant options.  Same for the other two circular trig buttons and their functions.  (See note below for a wish.)
• Holding the angle button reveals options for radian measures and DMS modes.
• Importantly, the comma key shows options for _ (underscore–for units), | (the constraint or substitution character, critical for many solving and CAS features), and quotation marks.
• Holding the parentheses keys reveals options for brackets and braces.

In short, there are MANY nice hidden options.  I’m very impressed and completely withdraw my earlier concern.  But …

NEW WISH:  I did notice that hyperbolic trig functions aren’t included anywhere, and it isn’t easy (as far as I can tell) without navigating keyboards to add an “h” to the end of one of the trig functions to create the hyperbolics.  I know I’m being picky and most high school math teachers don’t touch hyperbolic trigonometry, but it sure would be nice if either the trig buttons, or perhaps the $e^x$ button revealed some hyperbolic function options.

My concern about having to navigate between keyboards for necessary characters was unwarranted.  TI had already incorporated the fix.  Nice.

Thanks, Mr. Schirles!