# Tag Archives: quadratic

## Quadratics + Tangent = ???

Here’s a very pretty problem I encountered on Twitter from Mike Lawler 1.5 months ago.

I’m late to the game replying to Mike’s post, but this problem is the most lovely combination of features of quadratic and trigonometric functions I’ve ever encountered in a single question, so I couldn’t resist.  This one is well worth the time for you to explore on your own before reading further.

My full thoughts and explorations follow.  I have landed on some nice insights and what I believe is an elegant solution (in Insight #5 below).  Leading up to that, I share the chronology of my investigations and thought processes.  As always, all feedback is welcome.

WARNING:  HINTS AND SOLUTIONS FOLLOW

Investigation  #1:

My first thoughts were influenced by spoilers posted as quick replies to Mike’s post.  The coefficients of the underlying quadratic, $A^2-9A+1=0$, say that the solutions to the quadratic sum to 9 and multiply to 1.  The product of 1 turned out to be critical, but I didn’t see just how central it was until I had explored further.  I didn’t immediately recognize the 9 as a red herring.

Basic trig experience (and a response spoiler) suggested the angle values for the tangent embedded in the quadratic weren’t common angles, so I jumped to Desmos first.  I knew the graph of the overall given equation would be ugly, so I initially solved the equation by graphing the quadratic, computing arctangents, and adding.

Insight #1:  A Curious Sum

The sum of the arctangent solutions was about 1.57…, a decimal form suspiciously suggesting a sum of $\pi/2$.  I wasn’t yet worried about all solutions in the required $[0,2\pi ]$ interval, but for whatever strange angles were determined by this equation, their sum was strangely pretty and succinct.  If this worked for a seemingly random sum of 9 for the tangent solutions, perhaps it would work for others.

Unfortunately, Desmos is not a CAS, so I turned to GeoGebra for more power.

Investigation #2:

In GeoGebra, I created a sketch to vary the linear coefficient of the quadratic and to dynamically calculate angle sums.  My procedure is noted at the end of this post.  You can play with my GeoGebra sketch here.

The x-coordinate of point G is the sum of the angles of the first two solutions of the tangent solutions.

Likewise, the x-coordinate of point H is the sum of the angles of all four angles of the tangent solutions required by the problem.

Insight #2:  The Angles are Irrelevant

By dragging the slider for the linear coefficient, the parabola’s intercepts changed, but as predicted in Insights #1, the angle sums (x-coordinates of points G & H) remained invariant under all Real values of points A & B.  The angle sum of points C & D seemed to be $\pi/2$ (point G), confirming Insight #1, while the angle sum of all four solutions in $[0,2\pi]$ remained $3\pi$ (point H), answering Mike’s question.

The invariance of the angle sums even while varying the underlying individual angles seemed compelling evidence that that this problem was richer than the posed version.

Insight #3:  But the Angles are bounded

The parabola didn’t always have Real solutions.  In fact, Real x-intercepts (and thereby Real angle solutions) happened iff the discriminant was non-negative:  $B^2-4AC=b^2-4*1*1 \ge 0$.  In other words, the sum of the first two positive angles solutions for $y=(tan(x))^2-b*tan(x)+1=0$ is $\pi/2$ iff $\left| b \right| \ge 2$, and the sum of the first four solutions is $3\pi$ under the same condition.  These results extend to the equalities at the endpoints iff the double solutions there are counted twice in the sums.  I am not convinced these facts extend to the complex angles resulting when $-2.

I knew the answer to the now extended problem, but I didn’t know why.  Even so, these solutions and the problem’s request for a SUM of angles provided the insights needed to understand WHY this worked; it was time to fully consider the product of the angles.

Insight #4:  Finally a proof

It was now clear that for $\left| b \right| \ge 2$ there were two Quadrant I angles whose tangents were equal to the x-intercepts of the quadratic.  If $x_1$ and $x_2$ are the quadratic zeros, then I needed to find the sum A+B where $tan(A)=x_1$ and $tan(B)=x_2$.

From the coefficients of the given quadratic, I knew $x_1+x_2=tan(A)+tan(B)=9$ and $x_1*x_2=tan(A)*tan(B)=1$.

Employing the tangent sum identity gave

$\displaystyle tan(A+B) = \frac{tan(A)+tan(B)}{1-tan(A)tan(B)} = \frac{9}{1-1}$

and this fraction is undefined, independent of the value of $x_1+x_2=tan(A)+tan(B)$ as suggested by Insight #2.  Because tan(A+B) is first undefined at $\pi/2$, the first solutions are $\displaystyle A+B=\frac{\pi}{2}$.

Insight #5:  Cofunctions reveal essence

The tangent identity was a cute touch, but I wanted something deeper, not just an interpretation of an algebraic result.  (I know this is uncharacteristic for my typically algebraic tendencies.)  The final key was in the implications of $tan(A)*tan(B)=1$.

This product meant the tangent solutions were reciprocals, and the reciprocal of tangent is cotangent, giving

$\displaystyle tan(A) = \frac{1}{tan(B)} = cot(B)$.

But cotangent is also the co-function–or complement function–of tangent which gave me

$tan(A) = cot(B) = tan \left( \frac{\pi}{2} - B \right)$.

Because tangent is monotonic over every cycle, the equivalence of the tangents implied the equivalence of their angles, so $A = \frac{\pi}{2} - B$, or $A+B = \frac{\pi}{2}$.  Using the Insights above, this means the sum of the solutions to the generalization of Mike’s given equation,

$(tan(x))^2+b*tan(x)+1=0$ for x in $[0,2\pi ]$ and any $\left| b \right| \ge 2$,

is always $3\pi$ with the fundamental reason for this in the definition of trigonometric functions and their co-functions.  QED

Insight #6:  Generalizing the Domain

The posed problem can be generalized further by recognizing the period of tangent: $\pi$.  That means the distance between successive corresponding solutions to the internal tangents of this problem is always $\pi$ each, as shown in the GeoGebra construction above.

Insights 4 & 5 proved the sum of the angles at points C & D was $\pi/2$.  Employing the periodicity of tangent,  the x-coordinate of $E = C+\pi$ and $F = D+\pi$, so the sum of the angles at points E & F is $\frac{\pi}{2} + 2 \pi$.

Extending the problem domain to $[0,3\pi ]$ would add $\frac{\pi}{2} + 4\pi$ more to the solution, and a domain of $[0,4\pi ]$ would add an additional $\frac{\pi}{2} + 6\pi$.  Pushing the domain to $[0,k\pi ]$ would give total sum

$\displaystyle \left( \frac{\pi}{2} \right) + \left( \frac{\pi}{2} +2\pi \right) + \left( \frac{\pi}{2} +4\pi \right) + \left( \frac{\pi}{2} +6\pi \right) + ... + \left( \frac{\pi}{2} +2(k-1)\pi \right)$

Combining terms gives a general formula for the sum of solutions for a problem domain of $[0,k\pi ]$

$\displaystyle k * \frac{\pi}{2} + \left( 2+4+6+...+2(k-1) \right) * \pi =$

$\displaystyle = k * \frac{\pi}{2} + (k)(k-1) \pi =$

$\displaystyle = \frac{\pi}{2} * k * (2k-1)$

For the first solutions in Quadrant I, $[0,\pi]$ means k=1, and the sum is $\displaystyle \frac{\pi}{2}*1*(2*1-1) = \frac{\pi}{2}$.

For the solutions in the problem Mike originally posed, $[0,2\pi]$ means k=2, and the sum is $\displaystyle \frac{\pi}{2}*2*(2*2-1) = 3\pi$.

I think that’s enough for one problem.

APPENDIX

My GeoGebra procedure for Investigation #2:

• Graph the quadratic with a slider for the linear coefficient, $y=x^2-b*x+1$.
• Label the x-intercepts A & B.
• The x-values of A & B are the outputs for tangent, so I reflected these over y=x to the y-axis to construct A’ and B’.
• Graph y=tan(x) and construct perpendiculars at A’ and B’ to determine the points of intersection with tangent–Points C, D, E, and F in the image below
• The x-intercepts of C, D, E, and F are the angles required by the problem.
• Since these can be points or vectors in Geogebra, I created point G by G=C+D.  The x-intercept of G is the angle sum of C & D.
• Likewise, the x-intercept of point H=C+D+E+F is the required angle sum.

## Quadratics and CAS

I’m returning to my ‘blog after a prolonged absence.  My next several posts will explore ideas I shared and learned at the USACAS-10 conference hosted at Hawken School last weekend.

Finding equations for quadratic functions has long been a staple of secondary mathematics.  Historically, students are given information about some points on the graph of the quadratic, and efficient students typically figure out which form of the equation to use.  This post from my Curious Quadratics a la CAS presentation explores a significant mindset change that evolves once computer algebra enters the learning environment.

HISTORIC BACKGROUND:

Students spend lots of time (too much?) learning how to manipulate algebraic expressions between polynomial forms.  Whether distributing, factoring, or completing the square, generations of students have spent weeks changing quadratic expressions between three common algebraic forms

Standard:  $y=a*x^2+b*x+c$

Factored:  $y=a*(x-x_1)(x-x_2)$

Vertex:  $y=a*(x-h)^2+k$

many times without ever really knowing why.  I finally grasped deeply the reason for this about 15 years ago in a presentation by Bernhard Kutzler of Austria.  Poorly paraphrasing Bernhard’s point, he said in more elegant phrasing,

We change algebraic forms of functions because different forms reveal different properties of the function and because no single form reveals everything about a function.

While any of what follows could be eventually derived from any of the three quadratic forms, in general the Standard Form explicitly gives the y-intercept, Factored Form states x-intercepts, and Vertex Form “reveals” the vertex (duh).  When working without electronic technology, students can gain efficiency by choosing to work with a quadratic form that blends well with given information.  To demonstrate this, here’s an example of the differences between non-tech and CAS approaches.

COMPARING APPROACHES:

For an example, determine all intercepts and the vertex of the parabola that passes through $(10, 210)$$(5, 40)$, and $(-2, -30)$.

NON-TECH:  Not knowing anything about the points, use Standard form, plug in all three points, and solve the resulting system.

$y=a*x^2+b*x+c$
$210 = 100a+10b+c$
$40 = 25a+5b+c$
$-30 = 4a-2b+c$

Use any approach you want to solve this 3×3 system to get $a=2$, $b=4$, and $c=-30$.

That immediately gives the y-intercept at -30.  Factoring to $y=2(x+5)(x-3)$ or using the Quadratic Formula reveals the x-intercepts at -5 and 3.  Completing the square or leveraging symmetry between the known x-intercepts gives the vertex at $(-1,-32)$.  Some less-confident students find all of the hinted-at manipulations in this paragraph burdensome or even daunting.

CAS APPROACH:  By declaring the form you want/need, you can directly get any information you require.  In the next three lines on my Nspire CAS, notice that the only difference in my commands is the form of the equation I want in the first part of the command.  Also notice my use of lists to simplify substitution of the given points.

The last line’s output gave two solutions only because I didn’t specify which of x1 and x2 was the larger x-intercept, so my Nspire gave me both.

The -30 y-intercept appears in the first output, the vertex in the second, and the x-intercepts in the third.  Any information is equally simple to obtain.

CONCLUSION:

In the end, it’s all about knowing what you want to find and how to ask questions of the tools you have available.  Understanding the algebra behind the solutions is important, but endless repetition of these tasks is not helpful, even though it may be easy to test.

Instead, focus on using what you know, explore for patterns, and ask good questions.  …And teach with a CAS!

## Best Algebra 2 Lab Ever

This post shares what I think is one of the best, inclusive, data-oriented labs for a second year algebra class.  This single experiment produces linear, quadratic, and exponential (and logarithmic) data from a lab my Algebra 2 students completed this past summer.  In that class, I assigned frequent labs where students gathered real data, determined models to fit that data, and analyzed goodness of the models’ fit to the data.   I believe in the importance of doing so much more than just writing an equation and moving on.

For kicks, I’ll derive an approximation for the coefficient of gravity at the end.

THE LAB:

On the way to school one morning last summer, I grabbed one of my daughters’ “almost fully inflated” kickballs and attached a TI CBR2 to my laptop and gathered (distance, time) data from bouncing the ball under the Motion Sensor.  NOTE:  TI’s CBR2 can connect directly to their Nspire and TI84 families of graphing calculators.  I typically use computer-based Nspire CAS software, so I connected the CBR via my laptop’s USB port.  It’s crazy easy to use.

One student held the CBR2 about 1.5-2 meters above the ground while another held the ball steady about 20 cm below the CBR2 sensor.  When the second student released the ball, a third clicked a button on my laptop to gather the data:  time every 0.05 seconds and height from the ground.  The graphed data is shown below.  In case you don’t have access to a CBR or other data gathering devices, I’ve uploaded my students’ data in this Excel file.

Remember, this is data was collected under far-from-ideal conditions.  I picked up a kickball my kids left outside on my way to class.  The sensor was handheld and likely wobbled some, and the ball was dropped on the well-worn carpet of our classroom floor.  It is also likely the ball did not remain perfectly under the sensor the entire time.  Even so, my students created a very pretty graph on their first try.

For further context, we did this lab in the middle of our quadratics unit that was preceded by a unit on linear functions and another on exponential and logarithmic functions.  So what can we learn from the bouncing ball data?

LINEAR 1:

While it is very unlikely that any of the recorded data points were precisely at maximums, they are close enough to create a nice linear pattern.

As the height of a ball above the ground helps determine the height of its next bounce (height before –> energy on impact –> height after), the eight ordered pairs (max height #n, max height #(n+1) ) from my students’ data are shown below

This looks very linear.  Fitting a linear regression and analyzing the residuals gives the following.

The data seems to be close to the line, and the residuals are relatively small, about evenly distributed above and below the line, and there is no apparent pattern to their distribution.  This confirms that the regression equation, $y=0.673x+0.000233$, is a good fit for the = height before bounce and = height after bounce data.

NOTE:  You could reasonably easily gather this data sans any technology.  Have teams of students release a ball from different measured heights while others carefully identify the rebound heights.

The coefficients also have meaning.  The 0.673 suggests that after each bounce, the ball rebounded to 67.3%, or 2/3, of its previous height–not bad for a ball plucked from a driveway that morning.  Also, the y-intercept, 0.000233, is essentially zero, suggesting that a ball released 0 meters from the ground would rebound to basically 0 meters above the ground.  That this isn’t exactly zero is a small measure of error in the experiment.

EXPONENTIAL:

Using the same idea, consider data of the form (x,y) = (bounce number, bounce height). the graph of the nine points from my students’ data is:

This could be power or exponential data–something you should confirm for yourself–but an exponential regression and its residuals show

While something of a pattern seems to exist, the other residual criteria are met, making the exponential regression a reasonably good model: $y = 0.972 \cdot (0.676)^x$.  That means bounce number 0, the initial release height from which the downward movement on the far left of the initial scatterplot can be seen, is 0.972 meters, and the constant multiplier is about 0.676.  This second number represents the percentage of height maintained from each previous bounce, and is therefore the percentage rebound.  Also note that this is essentially the same value as the slope from the previous linear example, confirming that the ball we used basically maintained slightly more than 2/3 of its height from one bounce to the next.

And you can get logarithms from these data if you use the equation to determine, for example, which bounces exceed 0.2 meters.

So, bounces 1-4 satisfy the requirement for exceeding 0.20 meters, as confirmed by the data.

A second way to invoke logarithms is to reverse the data.  Graphing x=height and y=bounce number will also produce the desired effect.

Each individual bounce looks like an inverted parabola.  If you remember a little physics, the moment after the ball leaves the ground after each bounce, it is essentially in free-fall, a situation defined by quadratic movement if you ignore air resistance–something we can safely assume given the very short duration of each bounce.

I had eight complete bounces I could use, but chose the first to have as many data points as possible to model.  As it was impossible to know whether the lowest point on each end of any data set came from the ball moving up or down, I omitted the first and last point in each set.  Using (x,y) = (time, height of first bounce) data, my students got:

What a pretty parabola.  Fitting a quadratic regression (or manually fitting one, if that’s more appropriate for your classes), I get:

Again, there’s maybe a slight pattern, but all but two points are will withing  0.1 of 1% of the model and are 1/2 above and 1/2 below.  The model, $y=-4.84x^2+4.60x-4.24$, could be interpreted in terms of the physics formula for an object in free fall, but I’ll postpone that for a moment.

LINEAR 2:

If your second year algebra class has explored common differences, your students could explore second common differences to confirm the quadratic nature of the data.  Other than the first two differences (far right column below), the second common difference of all data points is roughly 0.024.  This raises suspicions that my student’s hand holding the CBR2 may have wiggled during the data collection.

Since the second common differences are roughly constant, the original data must have been quadratic, and the first common differences linear. As a small variation for each consecutive pair of (time, height) points, I had my students graph (x,y) = (x midpoint, slope between two points):

If you get the common difference discussion, the linearity of this graph is not surprising.  Despite those conversations, most of my students seem completely surprised by this pattern emerging from the quadratic data.  I guess they didn’t really “get” what common differences–or the closely related slope–meant until this point.

Other than the first three points, the model seems very strong.  The coefficients tell an even more interesting story.

GRAVITY:

The equation from the last linear regression is $y=4.55-9.61x$.  Since the data came from slope, the y-intercept, 4.55, is measured in m/sec.  That makes it the velocity of the ball at the moment (t=0) the ball left the ground.  Nice.

The slope of this line is -9.61.  As this is a slope, its units are the y-units over the x-units, or (m/sec)/(sec).  That is, meters per squared second.  And those are the units for gravity!  That means my students measured, hidden within their data, an approximation for coefficient of gravity by bouncing an outdoor ball on a well-worn carpet with a mildly wobbly hand holding a CBR2.  The gravitational constant at sea-level on Earth is about -9.807 m/sec^2.  That means, my students measurement error was about $\frac{9.807-9.610}{9.807}=2.801%$.  And 2.8% is not a bad measurement for a very unscientific setting!

CONCLUSION:

Whenever I teach second year algebra classes, I find it extremely valuable to have students gather real data whenever possible and with every new function, determine models to fit their data, and analyze the goodness of the model’s fit to the data.  In addition to these activities just being good mathematics explorations, I believe they do an excellent job exposing students to a few topics often underrepresented in many secondary math classes:  numerical representations and methods, experimentation, and introduction to statistics.  Hopefully some of the ideas shared here will inspire you to help your students experience more.

## CAS Presentations at USACAS-9

I had two presentations at last Saturday’s USACAS-9 conference at Hawken School in Cleveland, OH.  Following are outline descriptions of the two sessions with links to the PowerPoint, pdf, and .tns files I used.  I’m also adding all of this information to the Conference Presentations tab of this ‘blog.

Powerful Student Proofs

This session started with a brief introduction to a lab that first caught my eye at the first USACAS conference years ago.

You know how the graph of $y=ax^2+bx+c$ behaves when you vary a and c, but what happens when you change b?

I ‘blogged on this problem here and here.  In the session, we used TI-Nspire file QuadExplore.

Next, we explored briefly the same review of trigonometric and polar graphs not as static parent functions under static transformations, but as dynamic curves oscillating between their ceilings and floors.  In the session, we used TI-Nspire file Intro Polar.

Having a complete grasp of polar graphs of limacons, cardioids, rose curves, and hybrids of these, I investigated what would happen for curves of the family $r=cos \left( \frac{\theta}{k} \right)$.  Curiously, for $k=3$, I encountered a curve that looked like a horizontal translation of limacons–something that just shouldn’t happen within polar coordinates.

One of my former students, Sara, used a CAS to convert a polar curve to Cartesian, translate the curve, and convert back to polar.  She then identified and solved a trig identity to confirm what the graph suggested.  A complete description of Sara’s proof is below.  I originally ‘blogged on Sara’s work here which was a much more elegant solution to the problem than my initial attempt.  It’s always cool when a student’s work is better than her teacher’s!  I used TI-Nspire file Polar Fractions in Saturday’s session.

The last example presented itself when I created a document to model the family of conic curves resulting from manipulating the coefficients of $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$.  After I created  dynamic points for the foci of the conics, something unusual happened when the E parameter for horizontal ellipses and hyperbolas varied.

The foci for hyperbolas followed an ellipse, and the locus of elliptical foci appeared to be a hyperbola.  Another former student, Lilly, proved this property to be true.  A detailed explanation of Lilly’s proof is below.  We were fortunate to have Lilly’s work published in the Mathematics Teacher in May, 2014.

To demonstrate this final part of the session, I used TI-Nspire file Hidden Conic Behavior.

Here is my PowerPoint file for Powerful Student Proofs.  A more detailed sketch of the session and the student proofs is below.

Bending Asymptotes, Bouncing Off Infinity, and Going Beyond

The basic proposal was that adding the Reciprocal transformation to the palette of constant dilations and translations dramatically simplified understanding of the behavior of rational functions around even and odd vertical asymptotes (bouncing off and passing through infinity).  Just like lead coefficients of polynomials determine their end behavior, so, too, do the lead coefficients of proper rational expressions define the end behavior of rational functions.

Extending the idea of reciprocating and transforming functions, you can quickly explain exponential decay from exponential growth, derive the graphs of $y=\frac{1}{x}$ and $y=\frac{1}{x^2}$, and completely explain why logistic functions behave the way they do.

We finished with a quick exploration of trigonometric and polar graphs not as static parent functions under static transformations, but as dynamic curves oscillating between their ceilings and floors.

I used TI-Nspire Bending and Intro Polar files in the demonstration.  Here is my outline PowerPoint file for Bending Asymptotes.

## Student Quadratic Creativity

I’m teaching Algebra 2 this summer for my school.  In a recent test on quadratic functions, I gave a question I thought would be a little different, but still reachable for those willing to make connections or exert a little creativity.

Write a system of quadratic functions that has exactly one solution:  (1,1).

Their handheld graphing calculators were allowed.  Some students definitely had difficulty with the challenge, some gave a version of the answer I expected, and one adopted a form I knew was possible, but doubted anyone would actually find during a test situation.

I show my students’ solutions below.  But before you read on, can you give your own solution?

SOLUTION ALERT!  Don’t read further if you want to find your own solution.

WHAT I EXPECTED

We’ve had many discussions in class about the power of the Rule of 4–that math ideas can be expressed numerically, graphically, algebraically, and verbally.  When you get stumped in one representation, being able to shift to a different form is often helpful.  That could mean a different algebraic representation, or a different Rule of 4 representation altogether.

The question is phrased verbally asking for an algebraic answer.  But it asks about a solution to a system of equations.  I hoped my students would recall that the graphical version of a system solution is equivalent to the point(s) where the graphs of the equations intersected.  In my mind, the easiest way to do this is to write quadratic functions with coincident vertices.  And this is most easily done in vertex form.  The cleanest answer I ever got to this question was

A graphical representation verifies the solution.

Another student recognized that if two parabolas shared a vertex, but had different “slopes”, their only possible point of intersection was exactly the one the question required.  Here’s a graphical version of her answer.

From these two, you can see that there is actually an infinite number of correct solutions.  And I was asking them for just one of these!  🙂

WHAT I KNEW, BUT DIDN’T EXPECT

Another way to solve this question makes use of the geometry of quadratic graphs.  If two quadratics have the same leading coefficients, they are the same graph, intersect exactly once, or never intersect.  This is a very non-trivial idea for most students.  While I’m not convinced the author of the following solution had this in mind when he answered the question, his solution works because of that fact.  Here’s what J wrote on last week’s test and its graph.

J used more equations than he needed, but had he restricted himself to just two equations, I’m not sure the lovely pattern would have been so obvious.

This is a very different (and super cool) answer than what I expected my students to produce.  Lesson re-learned:  Challenge your students, give them room to express creativity and individuality, and be prepared to be amazed by them.

NEXT STEPS

J’s answer actually opens the door to other avenues of exploration.

1. Can you generalize the form of all of J’s equations, essentially defining a family of quadratics?  Can you prove that all members of your generalization satisfy the question posed and that no other answers are possible?
2. Can you find forms of other generalized families of quadratic functions whose only solution is (1,1)?
3. Notice that there were two types of solutions above:  A) those with coincident vertices and different lead coefficients and B) those with identical lead coefficients and different vertices.  Are these the only types of quadratics that can answer this question?  That is, is there a system of quadratics with (1,1) as the only solution that have identical vertices and lead coefficients?  Could both be different and (1,1) be the only solution?
4. If I relax the requirement that the quadratics be functions, what other types of quadratics are possible?  [This could be a very nice calculus question!]

For my part, I’m returning to some of these questions this week to stretch and explore my student’s creativity and problem-solving.

I’d love to hear what you or your students discover.

## Traveling Dots, Parabolas, and Elegant Math

Toward the end of last week, I read a description a variation on a paper-folding strategy to create parabolas.  Paraphrased, it said:

1. On a piece of wax paper, use a pen to draw a line near one edge.  (I used a Sharpie on regular copy paper and got enough ink bleed that I’m convinced any standard copy or notebook paper will do.  I don’t think the expense of wax paper is required!)
2. All along the line, place additional dots 0.5 to 1 inch apart.
3. Finally, draw a point F between 0.5 and 2 inches from the line roughly along the midline of the paper toward the center of the paper.
4. Fold the paper over so one of the dots on line is on tope of point F.  Crease the paper along the fold and open the paper back up.
5. Repeat step 4 for every dot you drew in step 2.
6. All of the creases from steps 4 & 5 outline a curve.  Trace that curve to see a parabola.

I’d seen and done this before, I had too passively trusted that the procedure must have been true just because the resulting curve “looked like a parabola.”  I read the proof some time ago, but I consumed it too quickly and didn’t remember it when I was read the above procedure.  I shamefully admitted to myself that I was doing exactly what we insist our students NEVER do–blindly accepting a “truth” based on its appearance.  So I spent part of that afternoon thinking about how to understand completely what was going on here.

What follows is the chronological redevelopment of my chain of reasoning for this activity, hopefully showing others that the prettiest explanations rarely occur without effort, time, and refinement.  At the end of this post, I offer what I think is an even smoother version of the activity, freed from some of what I consider overly structured instructions above.

CONIC DEFINITION AND WHAT WASN’T OBVIOUS TO ME

A parabola is the locus of points equidistant from a given  point (focus) and line (directrix).

What makes the parabola interesting, in my opinion, is the interplay between the distance from a line (always perpendicular to some point C on the directrix) and the focus point (theoretically could point in any direction like a radius from a circle center).

What initially bothered me about the paper folding approach last week was that it focused entirely on perpendicular bisectors of the Focus-to-C segment (using the image above).  It was not immediately obvious to me at all that perpendicular bisectors of the Focus-to-C segment were 100% logically equivalent to the parabola’s definition.

SIMILARITY ADVANTAGES AND PEDAGOGY

I think I had two major advantages approaching this.

1. I knew without a doubt that all parabolas are similar (there is a one-to-one mapping between every single point on any parabola and every single point on any other parabola), so I didn’t need to prove lots of cases.  Instead, I focused on the simplest version of a parabola (from my perspective), knowing that whatever I proved from that example was true for all parabolas.
2. I am quite comfortable with my algebra, geometry, and technology skills.  Being able to wield a wide range of powerful exploration tools means I’m rarely intimidated by problems–even those I don’t initially understand.  I have the patience to persevere through lots of data and explorations until I find patterns and eventually solutions.

I love to understand ideas from multiple perspectives, so I rarely quit with my initial solution.  Perseverance helps me re-phrase ideas and exploring them from alternative perspectives until I find prettier ways of understanding.

In my opinion, it is precisely this willingness to play, persevere, and explore that formalized education is broadly failing to instill in students and teachers.  “What if?” is the most brilliant question, and the one we sadly forget to ask often enough.

ALGEBRAIC PROOF

While I’m comfortable handling math in almost any representation, my mind most often jumps to algebraic perspectives first.  My first inclination was a coordinate proof.

PROOF 1:  As all parabolas are similar, it was enough to use a single, upward facing parabola with its vertex at the origin.  I placed the focus at $(0,f)$, making the directrix the line $y=-f$.  If any point on the parabola was $(x_0,y_0)$, then a point C on the directrix was at $(x_0,-f)$.

From the parabola’s definition, the distance from the focus to P was identical to the length of CP:

$\sqrt{(x_0-0)^2-(y_0-f)^2}=y_0+f$

Squaring and combining common terms gives

$x_0 ^2+y_0 ^2-2y_0f+f^2=y_0 ^2+2y_0f+f^2$
$x_0 ^2=4fy$

But the construction above made lines (creases) on the perpendicular bisector of the focus-to-C segment.  This segment has midpoint $\displaystyle \left( \frac{x_0}{2},0 \right)$ and slope $\displaystyle -\frac{2f}{x_0}$, so an equation for its perpendicular bisector is $\displaystyle y=\frac{x_0}{2f} \left( x-\frac{x_0}{2} \right)$.

Finding the point of intersection of the perpendicular bisector with the parabola involves solving a system of equations.

$\displaystyle y=\frac{x_0}{2f} \left( x-\frac{x_0}{2} \right)=\frac{x^2}{4f}$
$\displaystyle \frac{1}{4f} \left( x^2-2x_0x+x_0 ^2 \right) =0$
$\displaystyle \frac{1}{4f} \left( x-x_0 \right) ^2 =0$

So the only point where the line and parabola meet is at $\displaystyle x=x_0$–the very same point named by the parabola’s definition.  QED

Proof 2:  All of this could have been brilliantly handled on a CAS to save time and avoid the manipulations.

Notice that the y-coordinate of the final solution line is the same $y_0$ from above.

MORE ELEGANT GEOMETRIC PROOFS

I had a proof, but the algebra seemed more than necessary.  Surely there was a cleaner approach.

In the image above, F is the focus, and I is a point on the parabola.  If D is the midpoint of $\overline{FC}$, can I conclude $\overline{ID} \perp \overline{FC}$, proving that the perpendicular bisector of $\overline{FC}$ always intersects the parabola?

PROOF 3:  The definition of the parabola gives $\overline{FI} \cong \overline{IC}$, and the midpoint gives $\overline{FD} \cong \overline{DC}$.  Because $\overline{ID}$ is self-congruent, $\Delta IDF \cong \Delta IDC$ by SSS, and corresponding parts make the supplementary $\angle IDF \cong \angle IDC$, so both must be right angles.  QED

PROOF 4:  Nice enough, but it still felt a little complicated.  I put the problem away to have dinner with my daughters and when I came back, I was able to see the construction not as two congruent triangles, but as the single isosceles $\Delta FIC$ with base $\overline{FC}$.  In isosceles triangles, altitudes and medians coincide, automatically making $\overline{ID}$ the perpendicular bisector of $\overline{FC}$.  QED

Admittedly, Proof 4 ultimately relies on the results of Proof 3, but the higher-level isosceles connection felt much more elegant.  I was satisfied.

TWO DYNAMIC GEOMETRY SOFTWARE VARIATIONS

Thinking how I could prompt students along this path, I first considered a trace on the perpendicular lines from the initial procedure above (actually tangent lines to the parabola) using to trace the parabolas.  A video is below, and the Geogebra file is here.

http://vimeo.com/89759785

It is a lovely approach, and I particularly love the way the parabola appears as a digital form of “string art.”  Still, I think it requires some additional thinking for users to believe the approach really does adhere to the parabola’s definition.

I created a second version allowing users to set the location of the focus on the positive y-axis and using  a slider to determine the distances and constructs the parabola through the definition of the parabola.  [In the GeoGebra worksheet (here), you can turn on the hidden circle and lines to see how I constructed it.]  A video shows the symmetric points traced out as you drag the distance slider.

A SIMPLIFIED PAPER PROCEDURE

Throughout this process, I realized that the location and spacing of the initial points on the directrix was irrelevant.  Creating the software versions of the problem helped me realize that if I could fold a point on the directrix to the focus, why not reverse the process and fold F to the directrix?  In fact, I could fold the paper so that F touched anywhere on the directrix and it would work.  So, here is the simplest version I could develop for the paper version.

1. Use a straightedge and a Sharpie or thin marker to draw a line near the edge of a piece of paper.
2. Place a point F roughly above the middle of the line toward the center of the paper.
3. Fold the paper over so point F is on the line from step 1 and crease the paper along the fold.
4. Open the paper back up and repeat step 3 several more times with F touching other parts of the step 1 line.
5. All of the creases from steps 3 & 4 outline a curve.  Trace that curve to see a parabola.

This procedure works because you can fold the focus onto the directrix anywhere you like and the resulting crease will be tangent to the parabola defined by the directrix and focus.  By allowing the focus to “Travel along the Directrix”, you create the parabola’s locus.  Quite elegant, I thought.

As I was playing with the different ways to create the parabola and thinking about the interplay between the two distances in the parabola’s definition, I wondered about the potential positions of the distance segments.

1. What is the shortest length of segment CP and where could it be located at that length?  What is the longest length of segment CP and where could it be located at that length?
2. Obviously, point C can be anywhere along the directrix.  While the focus-to-P segment is theoretically free to rotate in any direction, the parabola definition makes that seem not practically possible.  So, through what size angle is the focus-to-P segment practically able to rotate?
3. Assuming a horizontal directrix, what is the maximum slope the focus-to-P segment can achieve?
4. Can you develop a single solution to questions 2 and 3 that doesn’t require any computations or constructions?

CONCLUSIONS

I fully realize that none of this is new mathematics, but I enjoyed the walk through pure mathematics and the enjoyment of developing ever simpler and more elegant solutions to the problem.  In the end, I now have a deeper and richer understanding of parabolas, and that was certainly worth the journey.

## Fun with Series

Two days ago, one of my students (P) wandered into my room after school to share a problem he had encountered at the 2013 Walton MathFest, but didn’t know how to crack.  We found one solution.  I’d love to hear if anyone discovers a different approach.  Here’s our answer.

PROBLEM:  What is the sum of $\displaystyle \sum_{n=1}^{\infty} \left( \frac{n^2}{2^n} \right) = \frac{1^2}{2^1} + \frac{2^2}{2^2} + \frac{3^2}{3^3} + ...$ ?

Without the $n^2$, this would be a simple geometric series, but the quadratic and exponential terms can’t be combined in any way we knew, so the solution must require rewriting.  After some thought, we remembered that perfect squares can be found by adding odd integers.  I suggested rewriting the series as

where each column adds to the one of the terms in the original series.  Each row was now a geometric series which we knew how to sum.  That  meant we could rewrite the original series as

We had lost the quadratic term, but we still couldn’t sum the series with both a linear and an exponential term.  At this point, P asked if we could use the same approach to rewrite the series again.  Because the numerators were all odd numbers and each could be written as a sum of 1 and some number of 2s, we got

where each column now added to the one of the terms in our secondary series.  Each row was again a geometric series, allowing us to rewrite the secondary series as

Ignoring the first term, this was finally a single geometric series, and we had found the sum.

Does anyone have another way?

That was fun.  Thanks, P.