# Tag Archives: Nspire

## Define Your Own Math Rule

My friend, Knox S., introduced me to this problem.  According to a post on The Telegraph’s Education page, this was originally  posted on Facebook by Randall Jones.

The first line is fine by the standard rules of arithmetic, but as soon as you read the 2nd and 3rd lines, you know something is amiss.  What could be the output of line 4?

The Telegraph post above claims there are two answers.  Sadly, that post suggests there are only two solutions.  The reality is that there is an infinite number of correct answers.

I first share the two most commonly proffered solutions suggested by the Telegraph as the only answers.  I follow this with Knox’s clever use of an incremental number base.  Finally, I offer a more generalized approach to support my claim of many more solutions.

STANDARD SOLUTIONS

Consistent with the first three lines, the same rule to line 4 “proves” the answer is 40:

While nothing requires it, this approach is recursive.  I’ve not seen anyone say this, but the 40 approach requires the equations to appear in the given order.  If you give the equations in a different order, the rule is no longer consistent.  In particular, if you wanted a 5th line, what would it be?  There’s nothing clear about how to extend this solution.

• THE ANSWER IS 96:  Alternatively, you can multiply the two numbers on the left and add that product to the first number.  This procedure is consistent with the first three lines, so the solution to line 4 must be 96:

The nice thing about this approach is that the solution is explicit, not recursive.  What’s obviously counter-intuitive is why you would first multiply the given numbers, and then why you would add the result to the first number, not the second.  This approach is consistent with the given information, so it is valid.

Unlike the first solution, this multiplicative approach is not commutative.  By this rule, 1+4 yields 5, as shown, but 4+1 would be $4+(4*1)=8$.  Nothing in the problem statement required commutativity, so no worries.

Another good aspect of this algorithm is that the order of the equations is now irrelevant.  It applies no matter what numbers are “added” on the left side of the equation.  This is definitely more satisfying.

CHANGE THE NUMBER BASE

• THE ANSWER IS 201:  Knox noticed that if you changed the number base, you could find another legit pattern.  The first line is standard arithmetic, but how could the next lines be consistent, too?  You know 2+5 doesn’t give 12 in standard base-10 arithmetic, but if you use base-5, $2+5=7=1*5^1+2*5^0=12_5$.

Unfortunately, in base-5, line 1 would be $(1+4)_5=10_5$ and line 3 would be $(3+6)_5=14_5$, both inconsistent.  Knox’s cleverest move was to vary the number base.  The 3rd line is true in base-4; since the 1st line is true in any base larger than five, he found a consistent pattern by applying base-6 to line 1:

Following this pattern, the next line would be base-3, giving 201 as the answer:

The best part of Knox’s solution is that he maintains the addition integrity of the left side.  The down-side is that this approach works for only one more line.  Any 5th line would give a base-2 (binary) answer, and since base-1 does not exist, the problem would end there.

Knox’s approach also allows you to use any numbers you want for the left-hand sums.  But notice that answers depend on where you write the sum.  For example, if (2+5) was in any other line, you would not get 12.  In line 1, $(2+5)_6=11_6$, in line 3, you’d get $(2+5)_4=13_4$.

By now, you should see that any any rule could work so long as you are consistent.  Because standard arithmetic does not apply, solvers should feel free to invoke any functions or algorithms desired.  One way to do this is to think of each line as the inputs (left side) and output (right side) of a three-variable function.

• THE ANSWER IS 96:  One possible function is $z=f(x,y)=a*x^2+b*y^2+c$ for some values of a, b, and c that passes through (1,4,5), (2,5,12), and (3,6,21).  I used my TI-Nspire CAS to solve the resulting system:

That means if x and y are the given left-side numbers and z is the right-side answer, the equation $\frac{1}{3}*x+\frac{2}{3}*y-6=z$ satisfies the first three lines and the answer to line 4 is 96

• THE ANSWER IS $\displaystyle \frac{2574}{29}$:  If you can square the inputs, why not cube them?  That means another possible function is $z=f(x,y)=a*x^3+b*y^3+c$.  My CAS solution of the resulting system leads to the fractional answer:

The first three given equations essentially define three ordered triples–(1,4,5), (2,5,12), and (3,6,21)–so almost any equation you conceive with three unknown coefficients can be used to create a 3×3 system of equations.  The fractional solution for line 4 may not be as satisfying as any of the earlier approaches using only integers, but these last two examples make it clear that there should be an infinite number of solutions.

These last two solutions are especially nice because they are explicit and don’t depend on the order of the given information.  You can choose any two numbers to “add”, and the algorithms will work.

Notice also that all of these functions, except for Knox’s, are non-commutative.  No worries, the problem already broke free of standard rules in line 2.

ONE THAT DIDN’T WORK

The last two examples prove the existence of quadratic and cubic solutions, so why not a linear solution?  In other words, is there a 3D plane in the form $z=a*x+b*y+c$ containing the given points?

Unfortunately, the resulting 3×3 system didn’t solve. The determinant of the coefficient matrix is zero, suggesting an inconsistent or dependent system.  Upon further inspection, subtracting line 1 from line 2 in the planar system gives $a+b=7$.  Similarly, subtracting line 2 from line 3 gives $a+b=9$.  Since both can’t be simultaneously true, the system is inconsistent and has no solution.  It was worth the effort.

CONCLUSION

Since standard arithmetic didn’t apply after the first line and no other restrictions were in play, that opened the door to lots of creativity.  The many different solutions to this problem all hinge on finding some function–any function–that satisfied the first three lines.  Find one of these, and the last line is simple.  That some attempts won’t work is no hinderance.  Even when standard algorithms seem to apply, there is almost always the possibility of some creative twist when working with numerical sequences.

So, whenever you’re faced with a non-standard system, have fun, be creative, and develop something unexpected.

## Circle and Square

Here’s another great geometry + algebra problem, posed by Megan Schmidt and pitched by Justin Aion to some students in his Geometry class.

Following is the problem as Justin posed it yesterday.

Justin described the efforts of three of his students’ on his his ‘blog.  Following is my more generalized approach.  Don’t read further if you want to solve this problem for yourself!

My first instinct in any case like this is build it in a dynamic geometry package and play.  Using my TI-Nspire, without loss of generality, I graphed a circle centered at the origin, constructed a tangent segment at the bottom of the circle centered on the y-axis, and then used that segment to construct a square.  I recognized that the locus of the upper right corners of all such squares would form a line.

That made it clear to me that for any circle, there was a unique square that intersected the circle three times as Megan had posed.

Seeing this and revealing its algebraic bias, my mind conceived an algebraic solution.  Assuming the radius of the circle is R, the equation of my circle is $x^2+y^2=R^2$ making the lower y-intercept of the circle $(0,-R)$.  That made $y=2x-R$ the locus line containing the upper right corner of the square.

To find generic coordinates of the upper right corner of the square in terms of R, I just needed to solve the system of equations containing the circle and the line.  That’s easy enough to compute by hand if you can handle quadratic algebra.  That manipulation is not relevant right now, so my Nspire CAS’s version is:

The output confirms the two intersections are $(0,-R)$ and the unknown at $\displaystyle \left( \frac{4R}{5} , \frac{3R}{5} \right)$.

Because of the horizontal symmetry of the square with respect to the y-axis, the system solution shows that the generic length of the side of the square is $\displaystyle 2\left( \frac{4R}{5} \right) = \frac{8R}{5}$.  The circle’s y-intercept at $(0,-R)$ means the generic diameter of the circle is $2R$.

Therefore, the generic ratio of the circle’s diameter to the square’s side length is

$\displaystyle \frac{diameter}{side} = \frac{2R}{(8R)/5} = \frac{5}{4}$.

And this is independent of the circle’s radius!  The diameter of the circle is always $\frac{5}{4}$ of the square’s side.

CONCLUSION:

For Megan’s particular case with a side length of 20, that gives a circle diameter of 25, confirming Justin’s students’ solution.

Does anyone have a different approach?  I’m happy to compile and share all I get.

AN ASIDE:

While not necessary for the generalized solution, it was fun to see a 3-4-5 right triangle randomly appear in Quadrant 1.

## 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.

## 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.

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.

## 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.

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.

Fun.

TI finally converted its Nspire calculators to the iPad platform and through this weekend only in celebration of 25 years of Teachers Teaching with Technology, they’re offering both of their Nspire apps at $25 off their usual$29.99, or $4.99 each. This is a GREAT deal, especially considering everything the Nspire can do! Clicking on either of the images below will take you to a description page for that app. In my opinion, if you’re going to get one of these, I’d grab the CAS version. It does EVERYTHING the non-CAS version does plus great CAS tools. Why pay the same money for the non-CAS and get less? You aren’t required to use the CAS tools, but I’d rather have a tool and not need it than the other way around. If you read my ‘blog, though, you know I strongly advocate for CAS use for anyone exploring mathematics. Now, on to my brief review of the new apps. MY REVIEW: From my experimentations the last few days, this app appears to do EVERYTHING the corresponding handheld calculators can do. I wouldn’t be surprised if there are a few things the computer version can do that the app can’t, but I haven’t been able to find it yet. In a few places, I actually like the iPad app better than either the handheld or computer versions. Here are a few. • When you start the app, your home page shows all of the documents available that have been created on the app. It’s easy enough to navigate there on the handheld or computer, but it’s a nice touch (to me) to see all of my files easily arranged when I start up. • A BRILLIANT addition is the ability to export your working files to share with others. Using the standard export button common to all iPad apps with export features, you get the ability to share your current doc via email or iTunes. • The calculator history items can now be accessed using a simple tap instead of just arrow key or mouse navigation. • Personally, I find it much easier to access the menus and settings with conveniently located app buttons. I prefer having my tools available on a tap rather than buried in menus. A nice touch, from my perspective. • Moving objects is easy. I was easily able to graph $y=x$ and the generic $y=a\cdot x^2+b\cdot x+c$ with sliders for each parameter. It’s easy to drag the slider values, and after a brief tap-and-hold, a pop-up gives you an option to animate, change settings, move, or delete your slider. • Also notice on the left side of the three previous screens that you have thumbnails of your currently open windows. With a quick tap, you can quickly change between windows. • One of the best features of the Nspire has always been its ability to integrate multiple representations of mathematical ideas. That continues here. As I said, the app appears to be a fully-functional variation of the pre-existing handheld and computer versions. • The 3D-graphing option from a graphing page seems much easier to use on the iPad app. Being able to use my finger to rotate a graph the way I want just seems much more intuitive than using my mouse. As with the computer software, you can define your 3D surfaces and curves in Cartesian function form or parametrically. • A lovely touch on the iPad version is the ability to use finger pinch and spread maneuvers to zoom in and out on 2D and 3D graphs. Dragging your finger over a 2D graph easily repositions it. Combined, these options make it incredibly easy to obtain good graphing windows. For now, I see two drawbacks, but I can easily deal with both given the other advantages. 1. This concern has been resolved. See my response here. At the bottom of the 3rd screenshot above, you can see that variable x is available in the math entry keyboard, but variables y and t are not. You can easily grab a y through the alpha keyboard. It won’t matter for most, I guess, but entering parametric equations on a graph page and solving systems of equations on a calculator page both require flipping between multiple screens to get the variable names and math symbols. I get issues with space management, but making parametric equation entry and CAS use more difficult is a minor frustration. 2. I may not have looked hard enough, but I couldn’t find an easy way to adjust the computation scales for 3D graphs. I can change the graph scales, but I was not able to get my graph of $z=sin \left( x^2 + y^2 \right)$ to look any smoother. As I said, these are pretty minor flaws. CONCLUSION: It looks like strong, legitimate math middle and high school math-specific apps are finally entering the iPad market, and I know of others in development. TI’s Nspire apps are spectacular (and are even better if you can score one for the current deeply discounted price). ## Controlling graphs and a free online calculator When graphing functions with multiple local features, I often find myself wanting to explain a portion of the graph’s behavior independent of the rest of the graph. When I started teaching a couple decades ago, the processor on my TI-81 was slow enough that I could actually watch the pixels light up sequentially. I could see HOW the graph was formed. Today, processors obviously are much faster. I love the problem-solving power that has given my students and me, but I’ve sometimes missed being able to see function graphs as they develop. Below, I describe the origins of the graph control idea, how the control works, and then provide examples of polynomials with multiple roots, rational functions with multiple intercepts and/or vertical asymptotes, polar functions, parametric collision modeling, and graphing derivatives of given curves. BACKGROUND: A colleague and I were planning a rational function unit after school last week wanting to be able to create graphs in pieces so that we could discuss the effect of each local feature. In the past, we “rigged” calculator images by graphing the functions parametrically and controlling the input values of t. Clunky and static, but it gave us useful still shots. Nice enough, but we really wanted something dynamic. Because we had the use of sliders on our TI-nSpire software, on Geogebra, and on the Desmos calculator, the solution we sought was closer than we suspected. REALIZATION & WHY IT WORKS: Last week, we discovered that we could use $g(x)=\sqrt \frac{\left | x \right |}{x}$ to create what we wanted. The argument of the root is 1 for $x<0$, making $g(x)=1$. For $x>0$, the root’s argument is -1, making $g(x)=i$, a non-real number. Our insight was that multiplying any function $y=f(x)$ by an appropriate version of g wouldn’t change the output of f if the input to g is positive, but would make the product ungraphable due to complex values if the input to g is negative. If I make a slider for parameter a, then $g_2(x)=\sqrt \frac{\left | a-x \right |}{a-x}$ will have output 1 for all $x. That means for any function $y=f(x)$ with real outputs only, $y=f(x)\cdot g_2(x)$ will have real outputs (and a real graph) for $x only. Aha! Using a slider and $g_2$ would allow me to control the appearance of my graph from left to right. NOTE: While it’s still developing, I’ve become a big fan of the free online Desmos calculator after a recent presentation at the Global Math Department (join our 45-60 minute online meetings every Tuesday at 9PM ET!). I use Desmos for all of the following graphs in this post, but obviously any graphing software with slider capabilities would do. EXAMPLE 1: Graph $y=(x+2)^3x^2(x-1)$, a 6th degree polynomial whose end behavior is up for $\pm \infty$, “wiggles” through the x-axis at -2, then bounces off the origin, and finally passes through the x-axis at 1. Click here to access the Desmos graph that created the image above. You can then manipulate the slider to watch the graph wiggle through, then bounce off, and finally pass through the x-axis. EXAMPLE 2: Graph $y=\frac{(x+1)^2}{(x+2)(x-1)^2}$, a 6th degree polynomial whose end behavior is up for $\pm \infty$, “wiggles” through the x-axis at -2, then bounces off the origin, and finally passes through the x-axis at 1. Click here to access the Desmos graph above and control the creation of the rational function’s graph using a slider. EXAMPLE 3: I believe students understand polar graphing better when they see curves like the limacon $r=2+3cos(\theta )$ moving between its maximum and minimum circles. Controlling the slider also allows users to see the values of $\theta$ at which the limacon crosses the pole. Here is the Desmos graph for the graph below. EXAMPLE 4: Object A leaves (2,3) and travels south at 0.29 units/second. Object B leaves (-2,1) traveling east at 0.45 units/second. The intersection of their paths is (2,1), but which object arrives there first? Here is the live version. OK, I know this is an overly simplistic example, but you’ll get the idea of how the controlling slider works on a parametrically-defined function. The$latex \sqrt{\frac{\left | a-x \right |}{a-x}}\$ term only needs to be on one of parametric equations.  Another benefit of the slider approach is the ease with which users can identify the value of t (or time) when each particle reaches the point of intersection or their axes intercepts.  Obviously those values could be algebraically determined in this problem, but that isn’t always true, and this graphical-numeric approach always gives an alternative to algebraic techniques when investigating parametric functions.

ASIDE 1–Notice the ease of the Desmos notation for parametric graphs.  Enter [r,s] where r is the x-component of the parametric function and s is the y-component.  To graph a point, leave r and s as constants.  Easy.

EXAMPLE 5:  When teaching calculus, I always ask my students to sketch graphs of the derivatives of functions given in graphical forms.  I always create these graphs one part at a time.  As an example, this graph shows $y=x^3+2x^2$ and allows you to get its derivative gradually using a slider.

ASIDE 2–It is also very easy to enter derivatives of functions in the Desmos calculator.  Type “d/dx” before the function name or definition, and the derivative is accomplished.  Desmos is not a CAS, so I’m sure the software is computing derivatives numerically.  No matter.  Derivatives are easy to define and use here.

I’m hoping you find this technology tip as useful as I do.

## Quadratics, Statistics, Symmetry, and Tranformations

A problem I assigned my precalculus class this past Thursday ended up with multiple solutions by the time we finished.  Huzzah for student creativity!

The question:

Find equations for all polynomial functions, $y=f(x)$, of degree $\le 2$ for which $f(0)=f(1)=2$ and $f(3)=0$.

After they had worked on this (along with several variations on the theme), four very different ways of thinking about this problem emerged.  All were valid and even led to a lesson I hadn’t planned–proving that, even though they looked different algebraically, all were equivalent.  I present their approaches (and a few extras) in the order they were offered in our post-solving debriefing.

The commonality among the approaches was their recognition that 3 non-collinear points uniquely define a vertical parabola, so they didn’t need to worry about polynomials of degree 0 or 1.  (They haven’t yet heard about rotated curves that led to my earlier post on rotated quadratics.)

Solution 1–Regression:  Because only 3 points were given, a quadratic regression would derive a perfectly fitting quadratic equation.  Using their TI-Nspire CASs, they started by entering the 3 ordered pairs in a Lists&Spreadsheets window.  Most then went to a Calculator window to compute a quadratic regression.  Below, I show the same approach using a Data&Statistics window instead so I could see simultaneously the curve fit and the given points.

The decimals were easy enough to interpret, so even though they were presented in decimal form, these students reported $y=-\frac{1}{3}x^2+\frac{1}{3}x+2$.

For a couple seconds after this was presented, I honestly felt a little cheated.  I was hoping they would tap the geometric or algebraic properties of quadratics to get their equations.  But I then I remembered that I clearly hadn’t make that part of my instructions.  After my initial knee-jerk reaction, I realized this group of students had actually done exactly what I explicitly have been encouraging them to do: think freely and take advantage of every tool they have to find solutions.  Nothing in the problem statement suggested technology or regressions, so while I had intended a more geometric approach, I realized I actually owed these students some kudos for a very creative, insightful, and technology-based solution.  This and Solution 2 were the most frequently chosen approaches.

Solution 2–Systems:  Equations of quadratic functions are typically presented in standard, factored, or vertex form.  Since neither two zeros nor the vertex were explicitly given, the largest portion of the students used the standard form, $y=a\cdot x^2+b\cdot x+c$ to create a 3×3 system of equations.  Some solved this by hand, but most invoked a CAS solution.  Notice the elegance of the solve command they used, working from the generic polynomial equation that kept them from having to write all three equations, keeping their focus on the form of the equation they sought.

This created the same result as Solution 1, $y=-\frac{1}{3}x^2+\frac{1}{3}x+2$.

CAS Aside: No students offered these next two solutions, but I believe when using a CAS, it is important for users to remember that the machine typically does not care what output form you want.  The standard form is the only “algebraically simple” approach when setting up a solution by hand, but the availability of technology makes solving for any form equally accessible.

The next screen shows that the vertex and factored forms are just as easily derived as the standard form my students found in Solution 2.

I was surprised when the last line’s output wasn’t in vertex form, $y=-\frac{1}{3}\cdot \left ( x-\frac{1}{2} \right )^2+\frac{25}{12}$, but the coefficients in its expanded form clearly show the equivalence between this form and the standard forms derived in Solutions 1 and 2–a valuable connection.

Solution 3–Symmetry:  Two students said they noticed that $f(0)=f(1)=2$ guaranteed the vertex of the parabola occurred at $x=\frac{1}{2}$.  Because $f(3)=0$ defined one real root of the unknown quadratic, the parabola’s symmetry guaranteed another at $x=-2$, giving potential equation $y=a\cdot (x-3)(x+2)$.  They substituted the given (0,2) to solve for a, giving final equation $y=-\frac{1}{3}\cdot (x-3)(x+2)$ as confirmed by the CAS approach above.

Solution 4–Transformations:  One of the big lessons I repeat in every class I teach is this:

If you don’t like how a question is posed.  Change it!

Notice that two of the given points have the same y-coordinate.  If that y-coordinate had been 0 (instead of its given value, 2), a factored form would be simple.  Well, why not force them to be x-intercepts by translating all of the given points down 2 units?

The transformed data show x-intercepts at 0 and 1 with another ordered pair at $(3,-2)$.  From here, the factored form is easy:  $y=a\cdot (x-0)(x-1)$.  Substituting $(3,-2)$ gives $a=-\frac{1}{3}$ and the final equation is $y=-\frac{1}{3}\cdot (x-0)(x-1)$ .

Of course, this is an equation for the transformed points.  Sliding the result back up two units, $y=-\frac{1}{3}\cdot (x-0)(x-1)+2$, gives an equation for the given points.  Aside from its lead coefficient, this last equation looked very different from the other forms, but some quick expansion proved its equivalence.

Conclusion:  It would have been nice if someone had used the symmetry noted in Solution 3 to attempt a vertex-form answer via systems.  Given the vertex at $x=\frac{1}{2}$ with an unknown y-coordinate, a potential equation is $y=a\cdot \left ( x-\frac{1}{2} \right )^2+k$.  Substituting $(3,0)$ and either $(0,2)\text{ or }(1,2)$ creates a 2×2 system of linear equations, $\left\{\begin{matrix} 0=a\cdot \left ( 3-\frac{1}{2} \right )^2+k \\ 2=a\cdot \left ( 0-\frac{1}{2} \right )^2+k \end{matrix}\right.$.  From there, a by-hand or CAS solution would have been equally acceptable to me.

That the few alternative approaches I offered above weren’t used didn’t matter in the end.  My students were creative, followed their own instincts to find solutions that aligned with their thinking, and clearly appreciated the alternative ways their classmates used to find answers.  Creativity and individual expression reigned, while everyone broadened their understanding that there’s not just one way to do math.

It was a good day.

## 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.

## Inverses of curves

A fellow Atlanta-area teacher and TI technology buff, Dennis Wilson, started talking to me several months ago about the power of using a locus to describe mathematical relationships. My own exposure to loci had been admittedly limited until that point. When I heard “locus”, my mind immediately went to geometry and the conic sections which I had “learned” by their locus definitions, but I really don’t think I had understood fully how useful a locus was.

The example I give below shows how to construct the inverse to any function using TI-Nspire software. The presentation uses computer software, but easily can be replicated on handheld Nspires. What I particularly like about this approach is that it gives a very strong visual representation of what had largely been a purely algebraic maneuver –switch x & y, and solve for y (if possible). I had always understood inverses as transformations in which the input and output variables reversed roles and knew that was equivalent to reflecting a given curve over y=x, but this approach significantly enhanced my understanding of locus and definitely improved my visual conception of the construction of an inverse.

What I’ve found most amazing is that this approach handles inverses of any curve you can graph in the function menu of the Nspire.

Finding graphs of inverses of non-functions has always been a challenging problem for computers. For example, to find the graph of the inverse of the ellipse using a function graph on an Nspire CAS, you could use the zeros command. Typically, this provides the entire graph, but this inverse approach results in only half of the ellipse and the corresponding half of its inverse. Because the zeros command has limitations for higher order polynomials, I hoped the approach would work for parametrically defined curves. The image below shows that the Geometry Trace still gives the outline of the curve (probably enough for most), but the locus doesn’t work. (I suspect this is because I’m creating the inverse from _dependent_ x- and y-coordinates.)

The same restrictions (probably for the same dependence-independence reasons) apply to polar curves, but at least you can trace the inverse.

I need to think more, but for now, thank you, Dennis, for opening a door in my mind that should have been opened long ago.