Tag Archives: geogebra

Cover Article

I was pretty excited yesterday when the latest issue of NCTM’s Mathematics Teacher arrived in the mail and the cover story was an article I co-wrote with a former student who’s now at MIT.

The topic was the finding and proof of a cool interconnected property of the foci of hyperbolas and ellipses that I made years ago when setting up my TI-Nspire CAS to model conic sections via the polynomial definition.

Image5a

Image5b

After pitching the idea to teachers at professional conferences for a couple years with no response, I asked one of my 9th grade students if she’d be interested in a challenge.  Her eventual proof paralleled mine, and our work together enhanced and polished each other’s understanding and proofs.

While all of the initial work was done with the TI-Nspire CAS, we wrote the article using GeoGebra so that readers could freely access Web-based documents to explore the mathematics for themselves.

You can access the article on the NCTM site here.

While a few minor changes happened after it was created, here is a pre-publication proof of the article.

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Dynamic Linear Programming

My department is exploring the pros and cons of different technologies for use in teaching our classes. Two teachers shared ways to use Desmos and GeoGebra in lessons using inequalities on one day; we explored the same situation using the TI-Nspire in the following week’s meeting.  For this post, I’m assuming you are familiar with solving linear programming problems.  Some very nice technology-assisted exploration ideas are developed in the latter half of this post.

My goal is to show some cool ways we discovered to use technology to evaluate these types of problems and enhance student exploration.  Our insights follow the section considering two different approaches to graphing the feasible region.  For context, we used a dirt-biker linear programming problem from NCTM’s Illuminations Web Pages.

DirtBikes

Assuming x = the number of Riders built and = the number of Rovers built,  inequalities for this problem are

linear1

We also learn on page 7 of the Illuminations activity that Apu makes a $15 profit on each Rider and $30 per Rover.  That means an Optimization Equation for the problem is Profit=15x+30y.

GRAPHING THE FEASIBLE REGION:

Graphing all of the inequalities simultaneously determines the feasible region for the problem.  This can be done easily with all three technologies, but the Nspire requires solving the inequalities for y first.  Therefore, the remainder of this post compares the Desmos and GeoGebra solutions.  Because the Desmos solutions are easily accessible as Web pages and not separate files, further images will be from Desmos until the point where GeoGebra operates differently.

Both Desmos and GeoGebra can graph these inequalities from natural inputs–inputing math sentences as you would write them from the problem information:  without solving for a specific variable.  As with many more complicated linear programming problems, graphing all the constraints at once sometimes makes a visually complicated feasible region graph.

linear2

So, we decided to reverse all of our inequalities, effectively  shading the non-feasible region instead.  Any points that emerged unshaded were possible solutions to the Dirt Bike problem (image below, file here).  All three softwares shift properly between solid and dashed lines to show respective included and excluded boundaries.

linear3

Traditional Approach – I (as well as almost all teachers, I suspect) have traditionally done some hand-waving at this point to convince (or tell) students that while any ordered pair in the unshaded region or on its boundary (all are dashed) is a potential solution, any optimal solution occurs on the boundary of the feasible region.  Hopefully teachers ask students to plug ordered pairs from the feasible region into the Optimization Equation to show that the profit does vary depending on what is built (duh), and we hope they eventually discover (or memorize) that the maximum or minimum profit occurs on the edges–usually at a corner for the rigged setups of most linear programming problems in textbooks.  Thinking about this led to several lovely technology enhancements.

INSIGHT 1:  Vary a point.

During our first department meeting, I was suddenly dissatisfied with how I’d always introduced this idea to my classes.  That unease and our play with the Desmos’ simplicity of adding sliders led me to try graphing a random ordered pair.  I typed (a,b) on an input line, and Desmos asked if I wanted sliders for both variables.  Sure, I thought (image below, file here).

linear7

— See my ASIDE note below for a philosophical point on the creation of (a,b).
— GeoGebra and the Nspire require one additional step to create/insert sliders, but GeoGebra’s naming conventions led to a smoother presentation–see below.

BIG ADVANTAGE:  While the Illuminations problem we were using had convenient vertices, we realized that students could now drag (a,b) anywhere on the graph (especially along the boundaries and to vertices of the feasible region) to determine coordinates.  Establishing exact coordinates of those points still required plugging into equations and possibly solving systems of equations (a possible entry for CAS!).  However discovered, critical coordinates were suddenly much easier to identify in any linear programming question.

HUGE ADVANTAGE:  Now that the point was variably defined, the Optimization Equation could be, too!  Rewriting and entering the Optimation Equation as an expression in terms of a and b, I took advantage of Desmos being a calculator, not just a grapher.  Notice the profit value on the left of the image.

linear6

With this, users can drag (a,b) and see not only the coordinates of the point, but also the value of the profit at the point’s current location!  Check out the live version here to see how easily Desmos updates this value as you drag the point.

From this dynamic setup, I believe students now can learn several powerful ideas through experimentation that traditionally would have been told/memorized.

STUDENT DISCOVERIES:

  1. Drag (a,b) anywhere in the feasible region.  Not surprisingly, the profit’s value varies with (a,b)‘s location.
  2. The profit appears to be be constant along the edges.  Confirm this by dragging (a,b) steadily along any edge of the feasible region.
  3. While there are many values the profit could assume in the feasible region, some quick experimentation suggests that the largest and smallest profit values occur at the vertices of the feasible region.
  4. DEEPER:  While point 3 is true, many teachers and textbooks mistakenly proclaim that solutions occur only at vertices.  In fact, it is technically possible for a problem to have an infinite number optimal solutions.  This realization is discussed further in the CONCLUSION.

ASIDE:  I was initially surprised that the variable point on the Desmos graph was directly draggable.  From a purist’s perspective, this troubled me because the location of the point depends on the values of the sliders.  That said, I shouldn’t be able to move the point and change the values of its defining sliders.  Still, the simplicity of what I was able to do with the problem as a result of this quickly led me to forgive the two-way dependency relationships between Desmos’ sliders and the objects they define.

GEOGEBRA’S VERSION:

In some ways, this result was even easier to create on GeoGebra.  After graphing the feasible region, I selected the Point tool and clicked once on the graph.  Voila!  The variable point was fully defined.  This avoids the purist issue I raised in the ASIDE above.  As a bonus, the point was also named.

linear4Unlike Desmos, GeoGebra permits multi-character function names.  Defining Profit(x,y)=15x+30y and entering Profit(A) allowed me to see the profit value change as I dragged point A as I did in the Desmos solution. The Profit(A) value was dynamically computed in GeoGebra as a number value in its Algebra screen.  A live version of this construction is on GeoGebraTube here.

linear5

At first, I wasn’t sure if the last command–entering a single term into a multivariable term–would work, but since A was a multivariable point, GeoGebra nicely handled the transition.  Dragging A around the feasible region updated the current profit value just as easily as Desmos did.

INSIGHT 2:  Slide a line.

OK, this last point is really an adaptation of a technique I learned from some of my mentors when I started teaching years ago, but how I will use it in the future is much cleaner and more expedient.  I thought line slides were a commonly known technique for solving linear programming problems, but conversations with some of my colleagues have convinced me that not everyone knows the approach.

Recall that each point in the feasible region has its own profit value.  Instead of sliding a point to determine a profit, why not pick a particular profit and determine all points with that profit?  As an example, if you wanted to see all points that had a profit of $100, the Optimization Equation becomes Profit=100=15x+30y.  A graph of this line (in solid purple below) passes through the feasible region.  All points on this line within the feasible region are the values where Apu could build dirt bikes and get a profit of $100.  (Of course, only integer ordered pairs are realistic.)

linear8

You could replace the 100 in the equation with different values and repeat the investigation.  But if you’re thinking already about the dynamic power of the software, I hope you will have realized that you could define profit as a slider to scan through lots of different solutions with ease after you reset the slider’s bounds.  One instance is shown below; a live Desmos version is here.

linear9

Geogebra and the Nspire set up the same way except you must define their slider before you define the line.  Both allow you to define the slider as “profit” instead of just “p”.

CONCLUSIONS:

From here, hopefully it is easy to extend Student Discovery 3 from above.  By changing the P slider, you see a series of parallel lines (prove this!).  As the value of P grows, the line goes up in this Illuminations problem.  Through a little experimentation, it should be obvious that as P rises , the last time the profit line touches the feasible region will be at a vertex.  Experiment with the P slider here to convince yourself that the maximum profit for this problem is $165 at the point (x,y)=(3,4).  Apu should make 3 Riders and 4 Rovers to maximize profit.  Similarly (and obviously), Apu’s minimum profit is $0 at (x,y)=(0,0) by making no dirt bikes.

While not applicable in this particular problem, I hope you can see that if an edge of the feasible region for some linear programming problem was parallel to the line defined by the corresponding Optimization Equation, then all points along that edge potentially would be optimal solutions with the same Optimization Equation output.  This is the point I was trying to make in Student Discovery 4.

In the end, Desmos, GeoGebra, and the TI-Nspire all have the ability to create dynamic learning environments in which students can explore linear programming situations and their optimization solutions, albeit with slightly different syntax.  In the end, I believe these any of these approaches can make learning linear programming much more experimental and meaningful.

Two Squares, Two Triangles, and some Circles

Here’s another fun twist on another fun problem from the Five Triangles ‘blog.  A month ago, this was posted.

What I find cool about so many of the Five Triangles problems is that most permit multiple solutions.  I also like that several Five Triangles problems initially appear to not have enough information.  This one is no different until you consider the implications of the squares.

I’ve identified three unique ways to approach this problem.  I’d love to hear if any of you see any others.  Here are my solutions in the order I saw them.  The third is the shortest, but all offer unique insights.

Method 1: Law of Cosines

This solution goes far beyond the intended middle school focus of the problem, but it is what I saw first.  Sometimes, knowing more gives you additional insights.

Because DEF is a line and EF is a diagonal of a square, I know m\angle CEF=45^{\circ}, and therefore m\angle CED=135^{\circ}.  \Delta CEF is a 45-45-90 triangle with hypotenuse 6, so its leg, CE has measure \frac{6}{\sqrt{2}}=3\sqrt{2}.  Knowing two sides and an angle in \Delta DEC means I could apply the Law of Cosines.

DC^2 = 4^2 + (3\sqrt{2})^2 - 2\cdot (3\sqrt{2}) \cdot \cos(135^{\circ})=58

Because I’m looking for the area of ABCD,  and that is equivalent to DC^2, I don’t need to solve for the length of DC to know the area I seek is 58.

Method 2: Use Technology

I doubt many would want to solve using this approach, but if you don’t see (or know) trigonometry, you could build a solution from scratch if you are fluent with dynamic geometry software (GeoGebra, TI-Nspire, GSP).  My comfort with this made finding the solution via construction pretty straight-forward.

  1. Construct segment EF with fixed length 6.
  2. Build square CEGF with diagonal EF.  (This can be done several ways.  I was in a transformations mood, so I rotated EF 90^{\circ} to get the other endpoints.)
  3. Draw line EF  and then circle with radius 4 through point E.
  4. Mark point D as the intersection of circle and line EF outside CEGF .
  5. Draw a segment through points and C.  (The square of the length of CD is the answer, but I decided to go one more step.)
  6. Construct square ABCD with sides congruent to CD.  (Again, there are several ways to do this.  I left my construction marks visible in my construction below.)
  7. Compute the area of ABCD.

Here is my final GeoGebra construction.

Squares1

Method 3: The Pythagorean Theorem

Sometimes, changing a problem can make it much easier to solve.

As soon as I saw the problem, I forwarded it to some colleagues at my school.  Tatiana wrote back with a quick solution.  In the original image, draw diagonal, CG, of square CEGF. Because the diagonals of a square perpendicularly bisect each other, that creates right \Delta DHC with legs 3 and 7.  That means the square of the hypotenuse of \Delta DHC (and therefore the area of the square) can be found via the Pythagorean Theorem.

DC^2 = 7^2+3^2 = 58

Method 4: Coordinate Geometry

OK, I said three solutions, and perhaps this approach is completely redundant given the Pythagorean Theorem in the last approach, but you could also find a solution using coordinate geometry.

Because the diagonals of a square are perpendicular, you could construct ECFG with its center at the origin.  I placed point C at (0,3) and point E at (3,0).  That means point D is at (7,0), making the solution to the problem the square of the length of the segment from (0,3) to (7,0).  Obviously, that can be done with the Pythagorean Theorem, but in the image below, I computed number i in the upper left corner of this GeoGebra window as the square of the length of that segment.

Squares2

Fun.

Systems of lines

Here’s an interesting variation of a typical (MS) problem I found by following the Five Triangles ‘blog: http://fivetriangles.blogspot.com/2013/09/97-no-triangle.html .

(Note:  If you sign up on this or other ‘blogs, you can get lots of problems emailed to you every time they are added.)

INITIAL SOLUTION

I know this question can absolutely be solved without using technology, but when a colleague asked if it was appropriate to use technology here (my school is one-to-one with tablet laptops), I thought it would be cool to share with her the ease and power of Desmos.  You can enter the equations from the problem exactly as given (no need to solve for y), or you can set up a graph in advance for your students and email them a direct link to an already-started problem.

If you follow this link, you can see how I used a slider (a crazy-simple addition on Desmos) to help students discover the missing value of a.

DesmosSolve

FOLLOW-UP

I suggest in this case that playing with this problem graphically would grant insight for many students into the critical role (for this problem) of the intersection point of the two explicitly defined lines.  With or without technology support, you could then lead your students to determine the coordinates of that intersection point and thereby the value of a.

Keeping with my CAS theme, you could determine those coordinates using GeoGebra’s brand new CAS View:

GeogebraCAS

Substituting the now known values of x and y into the last equation in the problem gives the desired value of a.

NOTE:  I could have done the sliders in GeoGebra, too, but I wanted to show off the ease of my two favorite (and free!) online math tools.

CONCLUSION

Thoughts?  What other ideas or problems could be enhanced by a properly balanced use of technology?

As an extension to this particular problem, I’m now wondering about the area of triangle formed for any value of a.  I haven’t played with it yet, but it looks potentially interesting.  I see both tech and non-tech ways to approach it.

Rotating Parabolas

Here’s a fun problem that continues to grow that Nurfatimah Merchant and I included in our textbook .

How many uniquely defined curves can you find whose graphs contain the points (1,1), (6,-3), and (7,3)?

NOTE:  Some of the algebra below is very intimidating to those who aren’t pretty good friends with mathematical symbol-pushing.  If you like, you can skip to the brief video clip at the end of this post showing all solutions.

SOLUTION ALERT — Don’t read any further if you want to play with this problem yourself.

Many students instantly think of vertical parabolas of the form y=a\cdot x^2+b\cdot x+c, but when I presented this at the MMC meeting in Chicago tonight, two stellar Geometry teachers on the front row suggested circles.  Others correctly noted that there are infinitely many curves defined solely by those three points, but I’ll talk about that in a second.  Most students first think about vertical parabolas and circles.

From the generic equation for a vertical parabola, one could plug in the three given ordered pairs and create a 3×3 system of linear equations.  I argue that a CAS is a good call here, especially when it can give the quadratic equation in multiple forms with equal ease.

That’s lots of algebra, but it was all completed very quickly using my CAS, proving another technology advantage is that all forms of an equation are equally easy to compute on a CAS.  When people try to solve this problem by hand, they invariably use the standard form only because the algebraic manipulations in the other cases are unwieldy, at best.  Here’s an image of the curve using the factored form of the equation.

The circle was also easy to get using a CAS to drive the algebra.

This graph shows the circle and the vertical parabola together.

A horizontal parabola can be obtained just as easily, but I’ll leave that one for you to discover.

This past summer, I was prodded by a teacher at the institute Nurfatimah and I ran for Westminster’s Center for Teaching to find all rotated parabolas which contained those points.  To do so, I thought that if a parabola rotated some \theta units contained those points, then if I could rotate the points back \theta units, the corresponding parabola would be vertical, and I could solve it using my CAS as above.

Using a transformation matrix on my original 3 points, I found their rotation images and substituted these into the standard form of a vertical parabola, a\cdot x^2+b\cdot x+c as shown below.

Those bottom three lines definitely look intimidating, but for any constant value of \theta, each is just a linear equation in a, b, and c.  While my Nspire CAS couldn’t solve that system directly and WolframAlpha timed out, I could accomplish the algebra step-by step on the Nspire.  (Click here if you have TI-Nspire CAS software and want to see my algebra.)

Geogebra 4.2 Beta did accomplish the solution in a single line.  Its solution to each coefficient is shown here:

Substituting the expressions in terms of \theta for a, b, and c back into the standard quadratic, y=a\cdot x^2+b\cdot x+c and rotating the equation back into place created a generic equation for a rotatable parabola through the given three points.  I used this equation to define my rotatable parabola through the 3 points with a slider for \theta here on GeoGebraTube (here is the original GeoGebra document).  You don’t get to manipulate it, but the following vimeo clip shows all of the rotated parabolas for 0\leq\theta\leq 2\pi.

As a final bang for the presentation, one teacher in the audience wondered what it would look like if a trace was placed on the rotating parabolas.  Easily done.  Whether doing this in an Nspire CAS or on GeoGebra, right click the curve and select trace on.  Dragging the slider for the angle through all of its possible values creates the following graph.

Now that’s just pretty!  The enveloped triangle in the center is precisely the triangle circumscribed by the circle found above.  It’s straight edges are defined by the degenerate cases of the parabolas at the instances when the vertices were stretched to infinity.

As some in tonight’s audience pointed out, there are also infinitely many rotated circles and ellipses defined by these points.  Another wondered what would happen if a trace was placed on the rotating vertex instead of the entire parabola. On that last question, we’re pretty certain that the curve is some sort of “tri-perbola”–a ‘hyperbola’ with three branches–whose vertices are the three given points.  We don’t know an equation for it (yet), so that and other great extensions of this are now problems for another day.  I’d love to hear what others think or find in this problem.