# Tag Archives: proof

## Squares and Octagons, A compilation

My last post detailed my much-too-long trigonometric proof of why the octagon formed by connecting the midpoints and vertices of the edges of a square into an 8-pointed star is always 1/6 of the area of the original square.

My proof used trigonometry, and responses to the post on Twitter  and on my ‘blog showed many cool variations.  Dave Radcliffe thought it would be cool to have a compilation of all of the different approaches.  I offer that here in the order they were shared with me.

Method 1:  My use of trigonometry in a square.  See my original post.

Method 2:  Using medians in a rectangle from Tatiana Yudovina, a colleague at Hawken School.

Below, the Area(axb rectangle) = ab = 16 blue triangles, and
Area(octagon) = 4 blue triangles – 2 red deltas..

Now look at the two green, similar triangles.  They are similar with ratio 1/2, making

Area(red delta) = $\displaystyle \frac{b}{4} \cdot \frac{a}{6} = \frac{ab}{24}$, and

Area(blue triangle) = $\displaystyle \frac{1}{16} ab$

So, Area(octagon) = $\displaystyle 2 \frac{ab}{24}-4\frac {ab}{16}=\frac{1}{6}ab$.

QED

Method 3:  Using differences in triangle areas in a square (but easily extended to rectangles)from @Five_Triangles (‘blog here).

Full solution here.

Method 4:  Very clever shorter solution using triangle area similarity in a square also from @Five_Triangles (‘blog here).

Full second solution here.

Method 5:  Great option Using dilated kitesfrom Dave Radcliffe posting as @daveinstpaul.

Full pdf and proof here.

Method 6:  Use fact that triangle medians trisect each other from Mike Lawler posting as @mikeandallie.

Tweet of solution here.

Method 7:  Use a coordinate proof on a specific square from Steve Ingrassia, a colleague at Hawken School.  Not a quick proof like some of the geometric solutions, but it’s definitely different than the others.

If students know the formula for finding the area of any polygon using its coordinates, then they can prove this result very simply with nothing more than simple algebra 1 techniques.   No trig is required.

The area of polygon with vertices (in either clockwise or counterclockwise order, starting at any vertex) of $(x_1, y_1)$, $(x_2, y_2)$, …, $(x_n, y_n)$ is

$\displaystyle Area = \left| \frac{(x_1y_2-x_2y_1)+(x_2y_3-x_3y_2)+...+(x_{n-1}y_n-x_ny_{n-1})}{2} \right|$

Use a 2×2 square situated with vertices at (0,0), (0,2), (2,2), and (2,0).  Construct segments connecting each vertex with the midpoints of the sides of the square, and find the equations of the associated lines.

• L1 (connecting (0,0) and (2,1):    y = x/2
• L2 (connecting (0,0) and (1,2):   y=2x
• L3 (connecting (0,1) and (2,0):  y= -x/2 + 1
• L4 (connecting (0,1) and (2,2):  y= x/2 + 1
• L5 (connecting (0,2) and (1,0):  y = -2x + 2
• L6 (connecting (0,2) and (2,1):  y= -x/2 + 2
• L7 (connecting (1,2) and (2,0):  y = -2x + 4
• L8 (connecting (2,2) and (1,0):  y = 2x – 2

The 8 vertices of the octagon come at pairwise intersections of some of these lines, which can be found with simple substitution:

• Vertex 1 is at the intersection of L1 and L3:   (1, 1/2)
• Vertex 2 is at the intersection of L3 and L5:  (2/3, 2/3)
• Vertex 3 is at the intersection of L2 and L5:  (1/2, 1)
• Vertex 4 is at the intersection of L2 and L4:  (2/3, 4/3)
• Vertex 5 is at the intersection of L4 and L6:  (1, 3/2)
• Vertex 6 is at the intersection of L6 and L7:  (4/3, 4/3)
• Vertex 7 is at the intersection of L7 and L8:  (3/2, 1)
• Vertex 8 is at the intersection of L1 and L8:  (4/3, 2/3)

Using the coordinates of these 8 vertices in the formula for the area of the octagon, gives

$\displaystyle \frac{ \left| 1/3 +1/3+0+(-1/3)+(-2/3)+(-1/3)+0 \right|}{2} = \frac{2}{3}$

Since the area of the original square was 4, the area of the octagon is exactly 1/6th of the area of the square.

## Squares and Octagons

Following is a really fun problem Tom Reardon showed my department last May as he led us through some TI-Nspire CAS training.  Following the introduction of the problem, I offer a mea culpa, a proof, and an extension.

THE PROBLEM:

Take any square and construct midpoints on all four sides.
Connect the four midpoints and four vertices to create a continuous 8-pointed star as shown below.  The interior of the star is an octagon.  Construct this yourself using your choice of dynamic geometry software and vary the size of the square.

Compare the areas of the external square and the internal octagon.

You should find that the area of the original square is always 6 times the area of the octagon.

I thought that was pretty cool.  Then I started to play.

MINOR OBSERVATIONS:

Using my Nspire, I measured the sides of the octagon and found it to be equilateral.

As an extension of Tom’s original problem statement, I wondered if the constant square:octagon ratio occurred in any other quadrilaterals.  I found the external quadrilateral was also six times the area of the internal octagon for parallelograms, but not for any more general quadrilaterals.  Tapping my understanding of the quadrilateral hierarchy, that means the property also holds for rectangles and rhombi.

MEA CULPA:

Math teachers always warn students to never, ever assume what they haven’t proven.  Unfortunately, my initial exploration of this problem was significantly hampered by just such an assumption.  I obviously know better (and was reminded afterwards that Tom actually had told us that the octagon was not equiangular–but like many students, I hadn’t listened).   After creating the original octagon, measuring its sides and finding them all equivalent, I errantly assumed the octagon was regular.  That isn’t true.

That false assumption created flaws in my proof and generalizations.  I discovered my error when none of my proof attempts worked out, and I eventually threw everything out and started over.  I knew better than to assume.  But I persevered, discovered my error through back-tracking, and eventually overcame.  That’s what I really hope my students learn.

THE REAL PROOF:

Goal:  Prove that the area of the original square is always 6 times the area of the internal octagon.

Assume the side length of a given square is $2x$, making its area $4x^2$.

The octagon’s area obviously is more complicated.  While it is not regular, the square’s symmetry guarantees that it can be decomposed into four congruent kites in two different ways.  Kite AFGH below is one such kite.

Therefore, the area of the octagon is 4 times the area of AFGH.  One way to express the area of any kite is $\frac{1}{2}D_1\cdot D_2$, where $D_1$ and $D_2$ are the kite’s diagonals. If I can determine the lengths of $\overline{AG}$ and $\overline {FH}$, then I will know the area of AFGH and thereby the ratio of the area of the square to the area of the octagon.

The diagonals of every kite are perpendicular, and the diagonal between a kite’s vertices connecting its non-congruent sides is bisected by the kite’s other diagonal.  In terms of AFGH, that means $\overline{AG}$ is the perpendicular bisector of $\overline{FH}$.

The square and octagon are concentric at point A, and point E is the midpoint of $\overline{BC}$, so $\Delta BAC$ is isosceles with vertex A, and $\overline{AE}$ is the perpendicular bisector of $\overline{BC}$.

That makes right triangles $\Delta BEF \sim \Delta BCD$.  Because $\displaystyle BE=\frac{1}{2} BC$, similarity gives $\displaystyle AF=FE=\frac{1}{2} DC=\frac{x}{2}$.  I know one side of the kite.

Let point I be the intersection of the diagonals of AFGH.  $\Delta BEA$ is right isosceles, so $\Delta AIF$ is, too, with $m\angle{IAF}=45$ degrees.  With $\displaystyle AF=\frac{x}{2}$, the Pythagorean Theorem gives $\displaystyle IF=\frac{x}{2\sqrt{2}}$.  Point I is the midpoint of $\overline{FH}$, so $\displaystyle FH=\frac{x}{\sqrt{2}}$.  One kite diagonal is accomplished.

Construct $\overline{JF} \parallel \overline{BC}$.  Assuming degree angle measures, if $m\angle{FBC}=m\angle{FCB}=\theta$, then $m\angle{GFJ}=\theta$ and $m\angle{AFG}=90-\theta$.  Knowing two angles of $\Delta AGF$ gives the third:  $m\angle{AGF}=45+\theta$.

I need the length of the kite’s other diagonal, $\overline{AG}$, and the Law of Sines gives

$\displaystyle \frac{AG}{sin(90-\theta )}=\frac{\frac{x}{2}}{sin(45+\theta )}$, or

$\displaystyle AG=\frac{x \cdot sin(90-\theta )}{2sin(45+\theta )}$.

Expanding using cofunction and angle sum identities gives

$\displaystyle AG=\frac{x \cdot sin(90-\theta )}{2sin(45+\theta )}=\frac{x \cdot cos(\theta )}{2 \cdot \left( sin(45)cos(\theta ) +cos(45)sin( \theta) \right)}=\frac{x \cdot cos(\theta )}{\sqrt{2} \cdot \left( cos(\theta ) +sin( \theta) \right)}$

From right $\Delta BCD$, I also know $\displaystyle sin(\theta )=\frac{1}{\sqrt{5}}$ and $\displaystyle cos(\theta)=\frac{2}{\sqrt{5}}$.  Therefore, $\displaystyle AG=\frac{x\sqrt{2}}{3}$, and the kite’s second diagonal is now known.

So, the octagon’s area is four times the kite’s area, or

$\displaystyle 4\left( \frac{1}{2} D_1 \cdot D_2 \right) = 2FH \cdot AG = 2 \cdot \frac{x}{\sqrt{2}} \cdot \frac{x\sqrt{2}}{3} = \frac{2}{3}x^2$

Therefore, the ratio of the area of the square to the area of its octagon is

$\displaystyle \frac{area_{square}}{area_{octagon}} = \frac{4x^2}{\frac{2}{3}x^2}=6$.

QED

EXTENSIONS:

This was so nice, I reasoned that it couldn’t be an isolated result.

I have extended and proved that the result is true for other modulo-3 stars like the 8-pointed star in the square for any n-gon.  I’ll share that very soon in another post.

I proved the result above, but I wonder if it can be done without resorting to trigonometric identities.  Everything else is simple geometry.   I also wonder if there are other more elegant approaches.

Finally, I assume there are other constant ratios for other modulo stars inside larger n-gons, but I haven’t explored that idea.  Anyone?

## 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.

## Invariable Calculus Project II

As Rocky hinted in his comment to my last post, $\displaystyle f(x)=\frac{k}{x}$  also has the constant area property.  Following a lead from Cohen, et al’s Student Research Projects in Calculus, I discovered long ago that with nothing more than knowledge of the product rule, the quotient and/or chain rule, and a healthy dose of some patient algebra manipulations, students could actually determine all functions which have the property that right triangles formed by the x– and y-axes and tangent lines to those functions have constant area.  This morning, I discovered a nice CAS approach that makes the problem accessible to far more students.

This makes another great project for calculus students who’ve just learned algebraic rules for differentiation:

Determine an equation for any twice-differentiable function–$y=g(x)$–with the property that all tangent lines to g in Quadrant I, along with the x- and y-axes, form triangles of constant area.

(Very minor) HINT:  At some point in solving this problem, you’ll need to make and use some assumptions about the values of $a, g(a), g'(a)$, and $g''(a)$.

SOLUTION ALERT!  Don’t read further if you want to solve this problem for yourself.

Assumptions:  Let $(a,g(a))$ be any arbitrary point on $y=g(x)$ in Quadrant I.  This makes $a>0$ and $g(a)>0$.  I also know $g'(a)<0$ because otherwise both of the x– and y-intercepts of the tangent line would not be positive, making the triangle’s area negative.  Finally, if $g''(a)=0$, then g would be a linear function, and there would be only one triangle.  To keep the problem interesting, I’m going to assume $g''(a)\ne 0$.

Setting up:  We no longer have a specific function, so everything must be in generalities.  A generalized equation for a tangent line to any function $y=g(x)$ at $x=a$ is

$y-g(a)=g'(a)\cdot (x-a)$.

From here, the generalized x-intercept is $\displaystyle a-\frac{g(a)}{g'(a)}$, and the y-intercept is $g(a)-a\cdot g'(a)$.  [Side note, the x-intercept is also the same form used in Newton’s Method for root approximations, a connection I’ll make later in the term when I’m teaching AP Calculus.]  Combining the generalized intercepts, I can write a generic area formula.

$\displaystyle Area = \frac{1}{2} \cdot \left( a - \frac{g(a)}{g'(a)} \right) \cdot \left( g(a) - a \cdot g'(a) \right)$

Differentiating and Cleaning Up:  Whenever I’ve used this problem in the past, my students and I have always used paper & pencil techniques.  While I’m quite comfortable with my algebraic manipulation skills, the significant majority of my students struggle with this part of the assignment.  As I was writing this post today, I finally had a technology insight that I should have years ago, given my long interest in CAS.  So, I’ll show the next portion in two different approaches, first traditional, and then via CAS.

APPROACH 1 – TRADITIONAL PAPER & PENCIL:  Trying to keep some brevity in this expanding post, I skip a few algebra steps below while providing some guiding explanations.  Finding a common denominator in the Area equation and recognizing a common factor leads to

$\displaystyle Area(a) = - \frac{1}{2} \cdot \frac{\left( a \cdot g'(a) - g(a) \right) ^2}{g'(a)}$ .

Applying the quotient rule with respect to a gives

$\frac{d(Area(a))}{da} = -\frac{1}{2} \cdot \frac{g'(a)\cdot 2(a\cdot g'(a)-g(a))(1\cdot g'(a)+a\cdot g''(a) - g'(a))- (a\cdot g'(a)-g(a))^2\cdot g''(a)}{(g'(a))^2}$ .

Remember that I seek functions whose tangent lines create constant area triangles, so $\displaystyle \frac{d(Area(a))}{da} = 0$.  Using this on the left and canceling some terms on the right gives

$\displaystyle 0 = -\frac{1}{2} \cdot \frac{2a\cdot g'(a)\cdot g''(a)(a\cdot g'(a)-g(a))-(a\cdot g'(a)-g(a))^2 \cdot g''(a)}{(g'(a))^2}$ .

Pulling out common factors and cleaning up a little more turns this into a completely factored form.

$\displaystyle 0 = - \frac{1}{2} \cdot \frac{(a\cdot g'(a)-g(a))\cdot g''(a)\cdot (a\cdot g'(a)+g(a))}{(g'(a))^2}$

APPROACH 2 – CAS:  Because the algebra was going to be complicated enough to make viewing on a handheld calculator very difficult, I used my TI-nSpire CAS Computer software to tackle the problem.  In the image below, I defined the area function in line 1 and differentiated with respect to a in line 2.  Equating the final equation to 0 gives the last algebraic line above from APPROACH 1.

Notice that I did not need to define a differentiation technique or to manipulate the factoring.  The results on both lines automatically accomplish the factoring I worked so carefully to establish earlier in APPROACH 1.  This is a beautiful example of what I see as a central benefit of CAS:  Keeping users focused on the mathematics of the problem situation.

Some students might actually be curious about how the challenge of differentiating line 1 could end up as relatively “clean” as the result in line 2.  GOOD!  CAS also inspires creative thinking.

Other than the differentiation step, everything else in APPROACH 1 was simple algebra.  Complicated, perhaps, but simple.  In fact, I don’t think it’s mathematics at all; it’s algebraic arithmetic.  I’m not disparaging the work or the approach, but I see mathematics as pattern recognition and big thinking.  I think CAS is completely justified in this problem.

Applying the Zero Product Property:  Our initial assumptions clear the denominator because $g'(a)<0$.  Because $g''(a)\ne 0$, I can eliminate that term, too.  With a and $g(a)$ both positive and $g'(a)<0$, the $(a\cdot g'(a)-g(a))$ term must be negative and therefore can be eliminated.  That drops the initially complicated differential equation to

$\displaystyle 0 = a\cdot g'(a)+g(a)$.

Finally–the Solution:  Depending on how much your students know, this last equation can be solved three different ways:  A) recognizing differentiation rules, B) solving a separable differentiable equation, or C) using a CAS solver.  I typically assign this problem so early in a calculus course that they have no idea what a differential equation is, making  the first approach the only available technique.  But this is also a great problem to introduce after learning about separable DEs.

APPROACH A:  If you look carefully, you can recognize the right side as the result of the product rule applied to $a\cdot g(a)$.  (In my experience, most students need some time, encouragement, and occasionally some hints to “see” this.)  Because the product rule result equals zero, the original expression must have equalled a constant.  That means $a\cdot g(a) = C$ for any constant, C.  Solving gives $\displaystyle g(a)=\frac{C}{a}$.  That means Rocky’s suggested family of functions at the top of this post, $\displaystyle f(x)=\frac{k}{x}$ not only produces triangles of constant area, it’s the only family of functions that does!  Very cool!

APPROACH B:  Rewriting the result of the Zero Product Property simplification using xs and ys gives $\displaystyle 0=x\cdot \frac{dy}{dx} +y$.  The variables can be rearranged to give $\displaystyle -\frac{dx}{x}=\frac{dy}{y}$.  Integration gives $-ln(x)+ln(C)=ln(y)$ for any random constant, $ln(C)$.  Logarithm properties lead to $y=\displaystyle \frac{C}{x}$, as before.

APPROACH C:  While I like the pattern recognition insights from the previous two approaches, the solution can also be found using a CAS.

On the TI-nSpire, c1 represents any random constant, so the DE solver again gives the same results.

Conclusion:  No matter what approaches you take, this problem shows that the only functions that have the property of their tangent lines producing constant area triangles.

Shiny.

## Invariable Calculus Project

Here’s one of my favorite calculus projects.  I initially discovered it over 20 years ago in Cohen, et al’s superb Student Research Projects in Calculus.

For $x>0$, what is true about every triangle formed by the x- and y-axes and any tangent line to $\displaystyle y=\frac{1}{x}$ ?  Prove thy claim.

I’d love to say nothing more than that, but I usually don’t.  The problem sounds vague in its statement, but is pretty simple to solve.  The hidden property is a delightful surprise.  I encourage you to try it out for yourself before reading further.

I just assigned the problem to one of my classes of seniors.  The class is a one-semester introduction to calculus for primarily students who’ve never been in honors and largely aren’t enamored by mathematics.  Most take the class to get an introduction to statistics (fall) and calculus (spring) before likely taking a course in one of these two in college and–for most–never taking another math course.  With that background in mind, I’ve probably scaffolded this iteration of the problem more than I should.  Here’s the assignment I gave them this week.

WARNING!  Partial Solution Alert!  Don’t read further if you want to solve the problem for yourself.

I typically use this project early in my introduction to derivatives and walk students through a little review and data gathering to help them discover the surprising hidden property.  While I don’t expect my students to do this, my default approach to geometric-type problems is to use a dynamic geometry package.  The animation below shows what happens when I varied the point of tangency while tracking the base, height, and area of the resulting triangle.

Well, I hope that animation screams something.  The x– and y-intercepts are the base and height, respectively, of a right triangle.  While those intercepts obviously vary as the point of tangency changes, the area of the triangle always seems to be 4.  It never changes!  If you’ve any geometry sense, something like that just shouldn’t happen.  So, is this a universal property, or is my animation misleading or limited in some way?  That’s a good question, and it requires proof.  Can you prove this apparent property about tangent lines to $\displaystyle y=\frac{1}{x}$?

FINAL SOLUTION ALERT!  Don’t read further if you want to prove this property for yourself.

For $\displaystyle f(x)=\frac{1}{x}$, $\displaystyle\frac{d}{dx}\left(f(x)\right)=\frac{-1}{x^2}$, so an equation for the tangent line to f at any point $x=a$ is

$\displaystyle \left(y-\frac{1}{a}\right)=\frac{-1}{a^2}\left(x-a\right)$.

The x-intercept of this generic line is $\left(2a,0\right)$, and its y-intercept is $\displaystyle \left(0,\frac{2}{a}\right)$.  Therefore, the area of the triangle formed by the x-and y-axes and the tangent line to f at any point $x=a$ is

$\displaystyle Area=\frac{1}{2}\cdot base\cdot height = \frac{1}{2}\cdot 2a\cdot\frac{2}{a}=2$.

Cool!  The triangle’s area is always 2, completely independent of the point of tangency!

EXTENSION:

Are there any other functions that have a similar property, or is $\displaystyle y=\frac{1}{x}$ alone in the mathematical universe for having constant area triangles?  Well, that’s a problem for another post.

## Quadrilateral surprise … or not?

The first time I recall encountering this problem was about 10-15 years ago when I was reading David Wells’ Curious and Interesting Geometry.  Not having taught geometry for several years, I’d forgotten about it until I encountered it again last week in Paul Lockhart’s Measurement.

My statement of the problem:

Draw any quadrilateral with non-intersecting sides.  Place a point at the midpoint of each side.  Connect the four midpoints in clockwise order.  1) What shape does the resulting quadrilateral always assume?  2) How does the area of the new quadrilateral compare to the area of the original quadrilateral?  Of course, you need to prove thy claims.

Draw several different quadrilaterals in your investigation.  So long as the sides don’t overlap, nothing else matters.  The pattern will emerge.  I love the stunning and unexpected emergence of order.

Don’t read any further until you’ve played with this for yourself.  The joy of mathematical discovery is worth it!  Above all, give yourself and your students lots of time to explore.  Don’t be too quick to offer suggestions.

I suggest using TI-nSpire, Geogebra, Geometer’s Sketchpad, or some other dynamic geometry software to model this.  Using my TI, I was able to quickly explore an entire spectrum of results.  Following are two representative images.

No matter what type of non-overlapping quadrilateral you draw, a parallelogram always seems to emerge.  I thought a while through various approaches to discover an elegant way to prove this, and in the process discovered the area solution.  Find your own proof before reading further.

My great insight happened when I imagined a diagonal drawn in a quadrilateral, splitting the original into two triangles.  That insight reminded me of a cool triangle property:  In any triangle, if you connect the midpoints of two sides, the resulting segment is parallel to and half the length of the third side.

In any quadrilateral ABCD, let W, X, Y, and Z be the respective midpoints of segments AB, BC, CD, and AD.  Draw diagonal AC of ABCD, creating triangles ABC and ACD.

By the triangle property noted above, segments WX and ZY are each parallel to and half the length of WXYZ.  That is sufficient to establish that WXYZ is a parallelogram.  QED.

Two triangles sharing the same base doesn’t seem like a condition imposing lots of order, but that is just enough to lay the inevitable conditions for creating a highly structured parallelogram.  The emergence of a highly ordered parallelogram from a seemingly random quadrilateral was inevitable!  As one of my students said, “Math works.”

To establish the area condition, I offer a proof without words.

I’d love to hear how any of you approach the problem.  I’ll post any responses.

## Exploring Sequences and Lines

Here’s another favorite problem that could be used for any middle or high school students who’ve been exposed to both arithmetic sequences and linear equations.

There is a family of lines, $Ax+By+C=0$, for which $\left\{ A, B, C \right\}$ can be any arithmetic sequence.  What do all members of this family have in common?

As with most great math problems, the problem is simply stated and can be approached from many different perspectives.  In the end, this one isn’t that difficult to crack, but the solution (at least to me) is not intuitively obvious from the problem statement. If you’ve not seen this before, please try it before reading further.

WARNING:  Problem Solution Follows

MOST COMMON STUDENT APPROACH:  Given the problem’s phrasing, most begin by writing out a few possible equations using different arithmetic sequences, typically with positive common differences.  After a few tries, most get a little frustrated as no obvious pattern emerges from the equations.

It is this point that is typically the most difficult for me as a teacher.  I want to help anyone who struggles, but “help” often means allowing others the room to struggle and to find ways of solving problems on their own.  Only when a student or group gets really frustrated do I sometimes ask, “Are there any other ways you can look at this problem or at your work?”

Eventually, most decide to graph their equations to see if anything pops out.  The following graph was submitted this past semester by one of my precalculus students using the free online Desmos calculator.

Two lines intersecting in a point is common.  Three or more in a single point almost always indicates something interesting.  Seven lines through a single point is screaming for attention!  From this graph, all lines in this family apparently contain the point (1,-2).  That seems a bit random until you investigate further, but pattern recognition is just half of the battle for a mathematician.  Now that something interesting has been discovered, a mathematician wants to know if this is a fluke or something inherent to all members of the family.

NOTE:  All graphs my students have produced over the years have always contained specific equations.  I don’t know that it’s any more enlightening, but I did create on Desmos a slider version of the graphs of this family with one slider for the initial term (A) and a second for its common difference (d).

UNIQUE SOLUTION METHODS FROM STUDENTS:

All successful solutions eventually rewrite the sequence $\left\{ A,B,C \right\}$ to $\left\{ A,A+d,A+2d \right\}$ where d is the common difference for a generic arithmetic sequence with initial term A.

Method I: After discovering the common point of intersection, most plug it into the left side of the equation and simplify to get

$Ax+By+C=A\cdot 1+\left( A+d\right)\cdot -2+\left( A+2d \right)=0$.

Because the left side reduces to zero for all generic arithmetic sequences, $\left\{ A,A+d,A+2d \right\}$, (1,-2) must be common to all members of this family.

A few students aren’t comfortable dealing with 0=0, so these tend to plug in $x=1$ and solve for y to get $y=-2$, proving that the y-coordinate for $x=1$ for all members of this family is always -2.

Method II:  A few students every year work algebraically from properties of arithmetic sequences.  For any arithmetic sequence, $\left\{ A,B,C \right\}$, $\frac{A+C}{2}=B$.  This rewrites to $1\cdot A-2\cdot B+C=0$, so whenever $\left( x,y \right)=\left(1,-2 \right)$, then $Ax+By+C=0$ is a fundamental property of all arithmetic sequences.

Personally, I think this method gets closest to explaining why the point (1,-2) is the common characteristic of this family.

Method III:  This year, I had a student take an approach I’d never seen before.  She defined one arithmetic sequence as $\left\{ a,a+d,a+2d \right\}$ and another as $\left\{ m,m+n,m+2n \right\}$ for any real values of a,d,m, and n.  This leads to a system of equations: $a\cdot x+(a+d)\cdot y+(a+2d)=0$ and $m\cdot x+(m+n)\cdot y+(m+2n)=0$ .  If you have some younger students or if all the variables make you nervous, the solution is available via Wolfram Alpha.

Still, this system is pretty easy to manipulate.  If you multiply the first equation by m and the second by a, the x-terms will eliminate with subtraction, giving

$m\cdot((a+d)\cdot y+(a+2d))-a\cdot((m+n)\cdot y+(m+2n))=0$.

Solving for y shows that all of the coefficients simplify surprisingly easily.

$((ma+md)-(am+an))\cdot y=-(ma+2md)+(am+2an)$
$(md-an)\cdot y = -2\cdot (md-an)\Longrightarrow y=-2$

From here, determining $x=1$ is easy, proving the relationship.

SOLUTIONS and APPROACHES NOT YET OFFERED BY STUDENTS:

Approach A:  High school students don’t often think about ways to simplify problem situations, especially at the beginning of problems.  One way I did that for this problem in later class discussions was to recognize that it one of the terms in the arithmetic sequence was 0, you didn’t need to deal with nearly as many terms. For example, if your sequence was ${1,0,-1}$, the linear equation would be $x-1=0$.  Similarly, the sequence $\left\{ 01,2 \right\}$ leads to $y+2=0$.  Obviously, the only thing these two lines have in common is the point (1,-2).  A proof of the property must still be established, but this is one of the fastest ways I’ve seen to identify the central property.

Approach B:  A purely algebraic approach to this problem could redefine the arithmetic sequence as $\left\{ a,a+d,a+2d\right\}$as before, giving:

$a\cdot x+(a+d)\cdot y+(a+2d)=0$

Collecting like terms gives

$(x+y+1)\cdot a+(y+2)\cdot d=0$.

The values of a and d must remain as parameters to include all possible arithmetic sequences.  Because the equation always equals 0, the coefficients of a and d are both 0, making $y=-2$ (for the coefficient of d) and therefore $x=1$.

EXTENSION:

We once had a test question at the end of the unit containing this exercise.  Basically, it reminded students that they had discovered that all lines $Ax+By+C=0$, for which $\left\{ A, B, C \right\}$ can be any arithmetic sequence contained the point (1,-2).  It then asked for an equation of a family of linear functions using the same arithmetic $\left\{ A, B, C \right\}$ that all contained the point (1,2).

The two most common responses we’ve seen involve a reflection or a vertical translation.  (1,-2) can become (1,2) by reflecting over the x-axis, so making the y-values negative would do the trick:  $Ax-By+C=0$.  Similarly, (1,-2) can become (1,2) by translating up 4 units, giving $Ax+B(y-4)+C=0$.