## The infinite sum which equals -1/12

On 16 January, 1913, Srinivasa Ramanujan, a clerk living in Madras, India, sent a letter to Oxford professor Thomas Hardy, which contained many of his intuitive mathematical musings, one of which led to the conclusion:

$\displaystyle \sum_{n=1}^{\infty} n = 1 + 2 + 3 + 4 + 5 + ... = \frac{-1}{12}$

In normal mathematics, this bizarre result would be just that — a bizarre result. In high school math, as well as in most university math courses, a response of $\frac{-1}{12}$ would get a mark of zero. On the face of it, the series is clearly divergent, and there is no reason to doubt that the actual answer should be $\infty$.

What is also clear is that Ramanujan and Hardy were not dummies, having helped to revolutionize early 20th century mathematics in England, as well as making people begin to take notice that India has, for a long time, made significant advancements in mathematics. It is also important to note that this bizarre result is used today in string theory, in the context of the Riemann Zeta function.

$\displaystyle \zeta(z) = \sum_{n=1}^\infty \frac{1}{n^z} = \frac{1}{1^z}+\frac{1}{2^z}+\frac{1}{3^z}+...$

The difference is that the Zeta function works with complex numbers. The zeta function does not replace ordinary summation. Also, a finite sum still exists in the real numbers if their partial sums are shown to converge.

The zeta function comes from a well-known summation which goes something like:

$\displaystyle \frac{1}{1-z} = 1 + z + z^2 + z^3 + z^4 + ... ; -1 < z < 1$

The restriction on z is important if the series is to converge. When z wanders outside those boundaries, the series diverges, and the sytem breaks down, and we can’t find the sum. The zeta function, however, attempts to generalize this equation for any value of z.

The value of these sums have meaning in a particular mathematical context. It is as if you encountered the imaginary number $i = \sqrt{-1}$, and said that the result is meaningless, since square roots are supposed to only return positive numbers. But in math involving the complex plane, it is quite meaningful, and enjoys actual use in the design of integrated circuits, for instance.

To see how Ramanujan encountered -1/12 as the sum of 1 + 2 + 3 + …, you can do normal summation, pretending that the series we are about to encounter always converge (and none of the following do):

$\displaystyle \frac{1}{1 + 1} = 1 - 1 + 1 - 1 + 1 - 1 + ... = \frac{1}{2}$

As you calculate partial sums, you get 1, 0, 1, 0, 1, …, and this never settles on any single value. This series is divergent. The 1/2 “sum” can be seen as an average of 0 and 1, and this actually is called a Cesaro sum.

Now suppose we squared this:

$\displaystyle \left(\frac{1}{1 + 1}\right)^2 = (1 - 1 + 1 - 1 + 1 - 1 + ...)^2 = 1 - 2 + 3 - 4 + ... = \frac{1}{4}$

The sum 1 – 2 + 3 – 4 … is the product of  the square of $(1 - 1 + 1 - 1 + 1 ...)^2$, which can be seen by using a grid to help you square them:

 × 1 -1 1 -1 1 … 1 1 -1 1 -1 1 … -1 -1 1 -1 1 1 … 1 1 -1 1 -1 1 … -1 -1 1 -1 1 1 … 1 1 -1 1 -1 1 … … … … … … … …

Then, you may add numbers from that table belonging to the same diagonal. Above, they happen to be the same colour. The 1’s and -1’s in gray are incomplete diagonals due to the limitations of a finite table.

You get: 1 – 2 + 3 – 4 + 5 – … = $\displaystyle \left(\frac{1}{2}\right)^2 = \frac{1}{4}$, which we will refer to as $S_2$.

Now, let’s get back to the original “sum”: $S_1 = 1 + 2 + 3 + 4 + 5 + ... = -1/12$. We can re-generate $S_2$ using only $S_1$ as follows:

\begin{aligned} S_1 & = & 1 + & 2 + & 3 + & 4 + & 5 + ... \\ 4S_1 & = & & 4 +& & 8 + & ... \\ S_1 - 4S_1 & = & 1 - & 2 + & 3 - & 4 + & 5 - ... \end{aligned}

… and finally, using our finding that S2 = 1/4, we arrive at the original incredible sum:

\begin{aligned} -3S_1 & = & \frac{1}{4} \\ S_1 & = & \frac{-1}{12} \end{aligned}

The sum is verified by the Riemann Zeta function for $\zeta(-1)$.

The feeling I get when I see a result like this is like when reading the Douglas Adams book Hitchhiker’s Guide To The Galaxy, and then being told by their whizzbang computer that the answer to life, the universe and everything was 42. You feel no more enlightened than before you asked the question. But then that is because the answer is not understood. And for that matter, even the question may need some work.

Mathematicians such as Euler and Ramanujan had endeavoured to unlock the mysteries of these strange results.

We know that $\zeta(z)$ works best for z > 1. For all z < 1 (especially negative numbers), $\zeta(z)$ does not return a direct value, but instead the mathematician needs to base a conjecture of what the value could be, based on analytical continuation. It turns out, this is far from guesswork since, for any negative z in the set of complex numbers, there is always exactly one answer that analytic continuation ever offers. $\zeta(-1) = -1/12$ is the result when $z = -1$.

The zeta function also equals zero in strange places, such as when z is a negative even integer. The reader is reminded that the set of complex numbers is utterly vast, subsuming the entire set of real numbers as a subset of the complex numbers. This is because a complex number C is defined as $C = a + bi$, where a and b belong to the set of real numbers, and i, once again, is $\sqrt{-1}$. If you allow $b = 0$, then C is a real number.

According to Riemann’s Hypothesis, the other set of zeroes for the zeta function lie entirely in the set of complex numbers where $C = \frac{1}{2} + bi$, for some value of b. In other words, all other zeroes for the Riemann zeta function are of the form $\zeta\left(\frac{1}{2} + bi\right)$ for some real number b.

Now, this is a hypothesis, which hasn’t been proved wrong yet, but it also hasn’t been proved to be correct generally. The Clay Mathematics Institute has made this one of their millenium prize problems, offering one million dollars to anyone who can solve it. If this is proven, other connected conjectures are proven along the way, such as the Goldbach Conjecture, and the Twin Prime Conjecture.

The locations of the zeroes of the zeta function appear to bear some relationship to the number of primes in an arithmetic progression with a fixed difference k. Proving Riemann will also validate the zeta function as useful for studying prime numbers, as had been done by Leonard Euler.

## Update on Tex editors

Nearly three years ago, I wrote about a comparison of LaTeX editors. Soon after, I began to use a third editor which, if you are a latex expert, you almost certaintly would have heard about, and are probably in fact using TeXStudio, an editor that has been around for close to a decade, but never appeared to show up on Linux installation packages. The editors that showed up, at least for me, were LyX and TeXmacs.

TeXstudio, once I discovered it, I installed it everywhere I could: on my Windows 10 and 7 machines, on my Linux installations, and even on Cygwin, even though they already had a Windows installation. To this day I have not seen any difference in output or functionality. All invocations of TeXstudio require a lot of time and packages for an installation of enough features.

First thing’s first: the editor. In LyX and TeXmacs, I needed to bail out of the editor, and export the code to LaTeX whenever I needed to do any serious equation editing or table editing or the like. In contrast, TeXstudio leaves me with no reason to ever leave the editor. First of all, the editor allows for native latex code to be entered. If there are pieces of Latex code that you don’t know, or have a fuzzy knowledge about, there is probably an icon or menu item that covers it. For document formatting, a menu item leads to a form dialog where you can fill in the form with sensible information pertaining to your particular document, default font size, paper size, margins, and so on. The ouput of this dialog is the preamble section to the LaTeX source file. To the rest of that source file, you add your document and formatting codes.  It is a kind of “notepad” for LaTeX, with syntax highlighting and shortcut buttons, menus and dialogs. It comes close to being WYSIWYG, in that “compiling” the code and pressing  the green “play” button brings up a window with the output of the existing code you are editing. It is not a live update, but it saves you the agony of saving, going on the command line compiling the code, and viewing in seeminly endless cycles. Now you can view the formatted document at the press of the play button.

## This is my first article in a while …

And I decided today to share what I learned about an algorithm for generating Pythagorean triples for any and , where Z.

A Pythagorean triple are any three whole numbers which satisfy the equation . For any two integers and , let ; ; and , and you will obtain a solution to the relation .

It is therefore not that hard, if we allow and to be any numbers from 1 to any upper limit you like, to write a computer program to generate the first Pythagorean triples, allowing for negative values for or .

## What the Golden Ratio has to do with The Thomas Crown Affair

There is a Geogebra app written by Ben Sparks which can run on your browser. There are many facets to this app, such as an animation of “circles within circles”. When this app runs (see the screenshot on your left), the dark points appear to move together in circles, but are really moving in straight line segments along the green paths. The red point A is the only point actually moving in a circle. The number of dark points can be controlled. When reduced to two points, the positions of them resemble the sine and cosine of angles if the red point was on the terminal arm of an angle in standard position. A circle of dark points can be formed from the points of intersection between a perpendicular passing through A and any line.

There was another one to do with a Mandlebrot set and a complex coordinate plane, even offering a “painter” which colors the canvas from black to blue when the iteration forming the spiral goes from stable to unstable, or from a cohesive spiral to one whose points are scattered as you drag a point C with your mouse over the coordinate plane. The points converge on a single point at the origin.

The last and my favourite, is one called a “sunflower spiral”. Apart from checkboxes, a text box for numeric input, and a “start/stop” button to control the animation, there is another checkbox which says “show lyrics”. They are the lyrics to the 1968 hit song “The Windmills of Your Mind”, which first appeared as the theme to the movie The Thomas Crown Affair, and sung by Noel Harrisson, and won an Academy Award for best original song. I won’t bore you with the lengthy list of vocalists who have covered it since, but I can provide a link if you really want to know that and other details.

The idea here is, imagine a flower with many, many seeds, like a sunflower. As it distributes more and more seeds, it must do so from the centre, pushing the seeds it has already deployed, more and more to the outside. Seeds thrive when they are further apart, so that they may take advantage of greater food availability.

This app appears to mimic a flower which distributes its “seeds”, symbolized as small circles, based on the distance from the centre and the amount of rotation based on the desired amount of seeds per turn. The number of turns has to be a number between 0 and 1. So, if you want 5 seeds per turn, seeds are distributed every 1/5 of a turn, and so you enter r = 0.2 in the text box. However, this means that, rather than spirals forming, you get spokes, as in the illustration above. However if you let it run for a bit, it does a fine iteration to six decimals, and you begin to see spirals bending, disappearing and re-forming, as in the illustration to the left, with a number like 0.206270. This is still a rational number, since the decimal terminates.

Notice that the centre of the “windmill” still has “spokes”. That might still be because 0.206270 is still close to 1/5, or 0.2. The outer “seeds” seem to arrange themselves in such a way that any one of them are members of at least two kinds of spirals, criss-crossing each other in opposing directions.

$\pi$ is another number thought to be highly irrational. Not irrational enough to scatter the seeds, as they fall in an orderly fashion and too close together for both $\pi - 3$ (above left) and $1/\pi$ (above right). Strangely, the spiral on the right has 7 “spokes”, and the one on the left has 22 “spokes”. If you recall your high school teacher telling you that an approximation for $\pi$ is $\frac{22}{7}$, I am not sure if that has anything to do with it.

Another candidate is Euler’s constant, e. I tried $e - 2$ (right) and $1/e$ (left), and both looked somewhat more satisfying. They still had distinct spirals somewhere in the seed distribution. When I say that the spirals are distinct/not distinct, I mean that the eye is drawn less and less to one particular spiral. But the seeds are overlapping a lot less, though there is still some overlap.

The irrational number $\sqrt{2}$ can be applied, and here, it is applied as $1/\sqrt{2}$ (left), and $\sqrt{2} - 1$ (right). This seems to be the best so far, as the spirals get less distinct, and the seeds appear to be spread further apart still, not seeming to touch anywhere.

For the golden ratio, this time the decimal was simply stripped from the number. Their button says $\frac{\sqrt{5}-1}{2}$, but this is the same as $\Phi - 1 = \frac{\sqrt{5}+1}{2}-1 = \frac{1}{\Phi}=0.61803398875$, oddly enough. The “seeds” appear to have the best distribution. I had the feeling this is true of any number $\frac{\sqrt{n}+1}{2}$, where n is a prime number, or at least a number that is not a perfect square. I had no luck, however, after trying 2, 3, 6, 7, and 11, and using the decimal or taking the reciprocal. Of course, the metallic means have numerical properties similar to the Golden Ratio, and two others were tried.

The reciprocal of the Silver Ratio, $\delta_S=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$ was already tried, and worked well (see above). The Bronze Ratio, $\frac{3+\sqrt{13}}{2}-3=0.302775638$, led to less satisfactory results.  Metallic means follow the formula ${\large\frac{n+\sqrt{n^2+4}}{2}}$, for any positive whole number n.

## Whatever happened to the conic sections: Parabolas

I don’t know about anywhere else, but in our school district, any parabola that is not solvable by the quadratic formula is an endangered species. Specifically, I am mourning the loss of parabolas whose directrix is not parallel to the x-axis. The ones that are parallel to the x-axis are such a mundane part of the curriculum that we never discuss the existence of a focus or a directrix.

Parabolas can open up in any direction, not just up or down. And they all possess a focus and a directrix. As an example, suppose the focus of the parabola was at the origin, and the directrix was the line $-x - y = 2$, or $y= -x-2$ if you prefer slope-intercept form. This ought to lead to a single parabola as a result: $x^2 - 2xy + y^2 -4x - 4y = 4$. Parabolas like these cannot be resolved to familiar equations of the form of $y= ax^2 + bx + c$, largely due to the difficulty in separating $x$ from $y$.

I have gotten used to saying to my students that all polynomial functions of $x$ have a domain in all real numbers. This works because I am only talking about functions. But for relations like $x^2 - 2xy + y^2 -4x - 4y = 4$, both domain and range have a restriction. The parabola looks like the illustration to the right. This relation is not a function, because  1) $x$ and $y$ can’t be separated, and 2) you can pass a vertical line through more than one point on its graph – that is, one value of $x$ generates two values of $y$. Other patterns we take for granted are also broken: the vertex is not a local extrema anymore. We require implicit differentiation to obtain the local extrema: $\frac{dy}{dx} = \frac{-x+y+2}{-x+y-2} = 0$ , which boils down to $y = x-2$. From the point of intersection between this line and the parabola, we find it will have an absolute minimum at $(1, -1)$. The range, then, is $y \ge -1$.

The vertex is located at $(-0.5, -0.5)$.

The domain is, similarly, $x \ge -1$.

## Happy π Day 2018!

Today, we mourn the passing of Stephen Hawking (age 76). He joins Howard Aiken, a computer designer from the early days, who died on $\pi$ day back in 1973. For the record, Albert Einstein was born this day back in 1879.

From 2014: A Mile of π, from Numberphile:

## The meaning of i^i

Imaginary numbers, like most things at the peripherals of our grasp of reality, interest me, since they appear to have their own rules at times. There was a video blog some time ago that tried to reckon with $\large i^i$ and considered what it equalled.

One calculation I observed came from De Moivre’s identity, $\large e^{i\theta} = \cos\theta + i \sin\theta$. If you let $\theta = \frac{\pi}{2}$, you get $e^{i\frac{\pi}{2}} = 0 + i = i$.

Now raising both sides of the equation $e^{i\frac{\pi}{2}} = i$ to the power i, you obtain the result: $e^{i\cdot i\frac{\pi}{2}} = i^i$, and, since $i\cdot i = i^2 = -1$, you get: $e^{\frac{-\pi}{2}} = i^i$.

This is a notable equation because the left hand side are all real numbers, and the right hand side are all complex. The result of the right-hand side is about 0.20788.

But did this crack that mystery? Well, if you consider the origin of the derivation was from De Moivre’s identity, then it would just compel one to think that not only is $\frac{\pi}{2}$ a solution, but all angles co-terminal to it. $i^i$ therefore has infinitely many answers if we are to rely on De Moivre’s Theorem.

## Representing the Quartic Formula

I have used various graphics packages to hunt for a quartic formula. Over the years I had settled for a formula that is one order down, the cubic formula. For a polynomial of the form $f(x) = ax^3 + bx^2 + cx + d$, all real solutions may be obtained by the formulae:

These formulae are relatively easy to find with a math package like Maple. See how the formula is quite a lot more complex when compared with the quadratic formula.

The formula for finding all real solutions to an order 4 polynomial has been elusive, however. Maple simply gives up and doesn’t bother. There is, however, Mathematica, which can come up with a quartic formula. I have a screenshot of all four solutions to $ax^4 + bx^3 + cx^2 + dx + e=0$, two parts at a time from the Mathematica output: (to view, you need to right click on the image and select “View image” or something similar on your browser)

The second part of this is:

Thank God Neils Abel verified that there is no such formula polynomials of order 5 and above. Another blogger went to the trouble of writing all four formulae in LaTeX, and came up with:

## 9×9 Magic Squares by the Lo Shu Method

The Lo Shu method generated the only 3×3 magic square that exists (rotations and reversals of the same 3×3 arrangement notwithstanding), and was the first magic square known to be discovered by humans, over 3000 years ago in China. Because of the one magic square it yields, utilizing the Lo Shu method recursively for 9×9 can only generate exactly one 9×9 magic square. A casual glance on Google will tell you that there are many more 9×9 squares out there that break the patterns that we are about to explore.

My preference has always been to discuss algorithms for magic squares that are “high-yielding”, or lead to many distinct magic squares, such as 5×5 (14,400 squares), and 7×7 (over 25 million squares). Why spend an article discussing the one possible square generated by the Lo Shu algorithm? I think this is because this has a great many unique patterns, and there are many “little” squares, each with their own magic numbers. It is not pan-magic in the normal way, but it seems to have multiple magical properties nonetheless.

The basic Lo Shu square begins with an ordered arrangement of the digits 1-9 in a 3×3 matrix:

                 1  2  3
4  5  6
7  8  9

Switch each corner number with the number kitty-corner to it:

                9  2  7
4  5  6
3  8  1

Then, imagine pressing the 9 to go between the 4 and 2, to create a new row, and doing the same with the 1 in placing it between the 8 and the 6. The diagonal 3  5  7 becomes the middle row of the new square. It would look thus:

                4  9  2
3  5  7
8  1  6

The result is a 3×3 square that is magic, and is the only 3×3 magic square of sequential digits 1-9 that exists. Beware of reflections and rotations of these squares, since they are still the same square.

The reason for showing the construction of the 3×3 square is because constructing the 9×9 square follows a strangely identical pattern.

### The 9×9 Lo Shu square is built on the same pattern

The same principle that created the 3×3 square can be used to make a 9×9 square. The drawback is that this method will again only yield one square. In truth, there are thousands of 9×9 squares possible. You begin with the numbers 1 to 81 in sequence in a 9×9 array:

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81

Let’s look at the first column. We make a 3×3 array out of this, starting by placing the numbers in numerical order in the same manner done with the numbers 1 to 9 earlier:

                         1   10   19
28   37   46
55   64   73

Then, we switch the corner numbers again:

                       73   10   55
28   37   46
19   64    1

Form new rows as described earlier, resulting in the top being 28  73  10, and the bottom being 64  1  46, with 19  37  55 being the middle row:

                       28   73   10
19   37   55
64    1   46

The result is a 3×3 sub-square that itself is magic. You can repeat this for the second column:

                       29   74   11
20   38   56
65    2   47

But then you would notice that each cell in this new 3×3 sub-square is one more than each corresponding cell in the previous 3×3 sub-square. Thus, the remaining sub-squares can be obtained by adding 1 to the previous sub-square when doing each of the columns in order from the original 9×9 square. You get the following intermediate 9×9 square:

 28 73 10 29 74 11 30 75 12 19 37 55 20 38 56 21 39 57 64 1 46 65 2 47 66 3 48 31 76 13 32 77 14 33 78 15 22 40 58 23 41 59 24 42 60 67 4 49 68 5 50 69 6 51 34 79 16 35 80 17 36 81 18 25 43 61 26 44 62 27 45 63 70 7 52 71 8 53 72 9 54

OK, so we are not quite there yet. The light grey and white patterns denote the nine 3×3 sub-squares that we were just discussing. But in the manner done for the initial 3×3 squares, we need to:

1. Switch out the corner 3×3 sub-squares, as shown below.
 36 81 18 29 74 11 34 79 16 27 45 63 20 38 56 25 43 61 72 9 54 65 2 47 70 7 52 31 76 13 32 77 14 33 78 15 22 40 58 23 41 59 24 42 60 67 4 49 68 5 50 69 6 51 30 75 12 35 80 17 28 73 10 21 39 57 26 44 62 19 37 55 66 3 48 71 8 53 64 1 46
2. Finally, make the magic square from the sub-squares going diagonally. That is, make a top row of the three sub-squares in the top left corner; then make a bottom row of the three sub-squares in the bottom right corner. The middle rows of the 9×9 square will consist of the three sub-squares extending from the bottom left to the top right. In this manner, we would have formed the magic square in the precise pattern that we used for individual numbers when we did the original 3×3 square. Below is the square, revelaling the magic number totals on the rows, columns, and diagonals:
 31 76 13 36 81 18 29 74 11 369 22 40 58 27 45 63 20 38 56 369 67 4 49 72 9 54 65 2 47 369 30 75 12 32 77 14 34 79 16 369 21 39 57 23 41 59 25 43 61 369 66 3 48 68 5 50 70 7 52 369 35 80 17 28 73 10 33 78 15 369 26 44 62 19 37 55 24 42 60 369 71 8 53 64 1 46 69 6 51 369 369 369 369 369 369 369 369 369 369 369 369

The feature of this square that intrigued me was in the way the algorithm was scalable from 3×3 to 9×9. That made it easy to learn.

## A Look at Geometry in Grade 10: Circles

Pedagogical content knowledge is a type of knowledge that is unique to teachers, and is based on the manner in which teachers relate their pedagogical knowledge (what they know about teaching) to their subject matter knowledge (what they know about what they teach) (Cochran, Kathryn, 1997). This concern seems especially pertinent for math teachers. I attempt to show how subject matter knowledge and pedagogical knowledge are both essential in getting students to the point where they can be assessed on a circles problem in grade 10 academic math.

Teachers are reminded at the outset that while knowing the subject is one essential ingredient, so is knowing your students and your age group. Students must be known individually, to become familiar with how they see the concepts, in their own words. This means that in a normal class setting, students must be able to be able to express themselves without fear of judgement. The teacher also needs to be familiar student IEPs, the supports in their school (student success teacher, guidance counsellors, social workers, and so on).
The teacher is urged to also try some geometry problems on their own. More than informing one’s content knowledge, the teacher is also learning to anticipate problems that may arise that affect lesson planning. According to Aslan-Tutak and Adams (2015), a lack of content knowledge robs teachers of being able to properly assess students real needs and strengths in the classroom.

This article takes a look at geometry in grade 10 and is written to raise awareness of the possible connectedness of aspects of geometry to other parts of the math program, and to also discuss implications for the learner.

In the Analytic Geometry strand, the Ministry (2005) gives as two of its overall expectations for the course MPM2D (Academic 10 Math):

1. By the end of the course, students will model and solve problems involving the intersection of two straight lines;
2. By the end of the course, students will solve problems using analytic geometry involving properties of lines and line segments.

Addressing both circles and the activity here involving them, the specific expectations look like this:

1. By the end of the course, students will develop the formula for the midpoint of a line segment, and use this formula to solve problems.
2. By the end of the course, students will  develop the formula for the equation of a circle with centre (0, 0) and radius r, by applying the distance formula for the length of  a line segment, $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$.
3. By the end of the course, students will determine the radius of a circle with centre (0, 0), given its equation; write the equation of a circle with centre (0, 0), given the radius; and sketch the circle, given the equation in the form $x^2 + y^2 = r^2$.

Grade 10 appears to be the first and last time circles are covered as a relation. I would also tell students about a circle centred at the point $(h, k)$, whose equation, $(x - h)^2 + (y - k)^2 = r^2$ is not that different from the distance formula shown above.

Student maturation. The chidren in this grade would generally be between 15-16 years old, and in most cases, maturationally ready to tackle math questions of some degree of complexity, even though impusivity is still an issue for most students at that age (Price, 2005). According to Price (2005), while impulsivity can be seen as a problem for adolescents of this age group regardless of their proficiency in math, she would also say that it can be regarded as an asset, that adolescent passion should be taken advantage of, and directed toward productive ends.

A task such as the three-point circle described below takes advatage of this passion, by subjecting a well-known property of circles to scrutiny. Beginning a rich task with a question starting with “Is it always true that three points always make a circle?” invites the student to try and make the idea fail. And of course, it does, sometimes. The student can then be asked, under what conditions does the idea fail, and can we re-state the conjecture that “three points make a circle” into something that is always true?

Big idea: Any three points can be used to form a circle if they are not on the same line. A rich task or rich assessment built on this will take with it a substatntial amount of the analytic geometry strand, and can wind up being among the last topics covered in a unit. The idea of three points making a circle is an old geometry problem. A much simpler, but less interesting, circle problem would be to have 3 points all some distance r from the origin to make a circle, which still loses none of the grade 10 content. Whenever I teach grade 10, I aim for a circle with an arbitrary centre, since so much of grade 10 math is embedded in it.

Rich problems such as this can be considered along the way as the result of a series of lessons. The sequence must be determined by the teacher

This can be either performed by the student using either geometry software, or using pencil and paper. It has been my experience that the latter option requires more time for the student, and more instructional time for the teacher, especially if the circle is not at the origin, which it likely won’t be given the prior requirement of “any three points”. This almost always requires the equation $(x - h)^2 + (y - k)^2 = r^2$, since the centre will be at some arbitrary point $(h, k)$ rather than the point $(0, 0)$.

It has been my experience that some learners take very well to this problem, while others are in need of assistance. If time is too tight, and students generally did well in the Quadratics strand, you might consider letting your students use Geometer’s Sketchpad to help solve the problem. Below, is a video where I demonstrate and discuss the use of Sketchpad for this problem.

Mostly, I will emphasize the benefits of a pencil-and-paper solution. And in the next video below, I demonstrate the use of an old fashioned geometry set and paper. The background music in this video is public domain, and there is no dialog, so feel free to turn off the volume (or turn it up). Since I was emphasizing technique in this video, I did not use graph paper.

A suggested lesson sequence to get to this point of the course (each of these could be one or more periods of lessons, covering other Big Ideas along the way):

1. Solving lines and linear systems — point of intersection: a) Solve by substitution; b) Solve by elimination (this could take several periods)
2. Midpoint of a line segment (can teach length of a line segment during this time to help confirm what we obtain from the formula $\left( \frac{x_2 - x_1}{2}, \frac{y_2 - y_1}{2}\right)$ is really the midpoint) (1-2 periods)
3. Solving quadratic systems (in Quadratic relations, the part on expansion of factors and solving is essential; however, the quadratic formula is not needed for this activity). (a couple of weeks)
4. The equation of a circle a) centred at the origin ($x^2 + y^2 = r^2$) and b) centred at $(h, k)$: $(x - h)^2 + (y - k)^2 = r^2$. Remind students of some properties of circles. For example, the circle is a collection of all points which are the same distance r from a centre point. (this can be 2-3 periods)
5. A look at chords, along with major and minor arcs (optional, but is helpful in establishing a terminology for the three point/circle problem). Don’t spend a lot of time on this — it is not in the Ministry, but it is in some Ministry-approved grade 10 texts for the current curriculum.
6. Students would need to play with trying to get three arbitrary points to make a circle for about 1 period to agree on what steps they would need to confirm the point/circle problem. They would either use software such as Sketchpad or a geometry set, but the decision must be made for the whole class. Students consolidate on what the steps ought to be to solve this problem. If software is used, I would add to the problem: find the full equation for the circle, its radius, and the centre point. If pencil and paper is used, after students have struggled, and an algorithm is decided upon, you might consider demonstrating a complete solution either on the board or using a document camera.
7. Once the class agrees to an algorithm, 1-2 periods would be spent on a rich task (possibly a summative). If pencil and paper is used, the question is still do-able by grade 10 students, but I find not everyone can identify the centre with algebra, and usually end up estimating the position of the centre from the graph drawn. Thus, the there would also be a loss of accuracy in computing radius. (I would give a level 4 for the algebra; level 3 for estimating).

For the latter topic, that is, the algebraic solution to finding the centre from three points on a circle, I would suggest a PDF I wrote for my students, which demonstrated a sample calculation as they were working on their problem, given to them, or demonstrated the some days before the assessment, especially if anyone is having problems. Not all steps are shown in this handout, and the student is advised to perform the steps themselves. They would also be given a worksheet to practice on. It is from the PDF above mentioned that students began to point out the resemblance between the general circle equation and the distance formula, because both are used in the activity.

Supporting the teaching and learning of mathematics.

Going though with the pencil-and-paper method teaches students several things which would not be seen on a computer system:

• Students see that there is now a broader use for expanding and solving quadratics, that isn’t part of the strand on quadratic relations, but uses techniques that are not foreign to it.
• It is one opportunity to prepare students for the kind of math they may encounter in grade 11 Functions, grade 12 Advanced Functions, and Calculus and Vectors.
• This is the last treatment students get with circle relations before university. It is no longer covered in grades 11 and 12, but such relations (and much more) are covered in first-year university texts.
• Students see that there are parallels that can now be drawn between solving for a linear system and solving for a quadratic system — you still need two equations with the same two unknowns, for one thing.
• What has preceeded shows that the unit must be very carefully planned to do this activity. But once done, the reward is an activity that captures a substantial part of grade 10 academic math.

Supporting high teacher efficacy.

In this page, I have covered most of the high points of this sort of lesson, and described in some detail much of the most difficult parts of it through videos and external documents.

In this page, I have told teachers how to prepare themselves and their students with suitable content knowledge, as well as what pertinent expectations are covered in the analytic geometry strand, as well as informing teachers of the maturational readiness of students, and how impusivity, re-directed as passion, can be used as an asset to student learning.

What I have not mentioned is that it is crucial that teachers must constantly assess their students in the “for” and “as” learning phases, through observations, conversations, as well as products. Most periods should not end without some kind of assessment of this nature. This is because, with this math, finding out where your students are in their learning is critical to understand next steps for planning. The consolidation phase could be a math congress where students share their findings, ask each other questions, and, with some guiding questions and information from the teacher, come to an agreement as to their general findings.

For teachers to be successful  in this strand, they would be best off with problem-based learning (PBL), using problems of varying degrees of open-endedness. This would be done for all or most lesson on the way to this one. PBL should be in the “Action” part of the 3-part lesson. Each lesson should not go without a consolidation phase (Ministry, 2010), where students share and explain their work, while answering questions.

Another aspect of geometry is covered by another video I did, one on quadrilaterals, with the question being: “Is it always true that midpoints on a quadrilateral make a parallellogram?” To see the answer, you have to accept rectangles and squares as special cases of parallellograms:

Bibliography/Resources

Aslan-Tutak, Fatma, and Thomasenia Adams. “A Study Of Geometry Content Knowledge Of Elementary Preservice Teachers.” International Electronic Journal Of Elementary Education, vol 7, no. 3, 2017, pp. 301-318.

Cochran, Kathryn (1997). Pedagogical Content Knowledge: Teachers’ Integration of Subject Matter, Pedagogy, Students, and Learning Environments [online] Available at: https://www.narst.org/publications/research/pck.cfm [Accessed 24 Jul. 2017].

Handal, Boris et al. (2013). “Technological Pedagogical Content Knowledge Of Secondary Mathematics Teachers – CITE Journal.” Citejournal.Org, http://www.citejournal.org/volume-13/issue-1-13/mathematics/technological-pedagogical-content-knowledge-of-secondary-mathematics-teachers/.

King, P. (2015). “Draw A Circle With Any Three Non-Collinear Points.” Youtube, 2015, https://youtu.be/ZdPQA6eSZD0.