# Tag Archives: math puzzle

## Pi-ling Up On Pi

Friends, Romans, countrymen, lend me your ears …

Oops, sorry, you can have your ears back (hope they still fit on the sides of your head); the ides of March are not ’til tomorrow.

But still, friends of pi, do pay heed! For there is foul news afoot this year on pi day (today, 3/14). The naysayers are clamoring against the number pi and its day.

A new majority in Congress thinks such celebrations are frivolous:

“Last Congress, Democrats voted hand over fist to increase spending, in addition to overusing the suspension calendar for hundreds of needless votes on resolutions honoring items such as Confucius’ birthday and ‘PI’ day,” wrote Brad Dayspring, a spokesman for House Majority Leader Eric Cantor (R-Va.)

Bloggers question the narrowness of confining ourselves to the decimal system and to the middle-endian calendar. [Yes, I’m aware the second blogger is me; I enjoy arguing against myself.]

And the new hero of the math vlogosphere, with her wonderful series about doodling in math class, Vi Hart, has even taken a stand against Pi and Pi Day. Instead she advocates for Tau Day (a.k.a 2pi day, on 6/28) since tau is “more natural” than pi. Click on that latter link and watch her anti-pi video, as it is our most formidable opponent, taking on not just the representation of pi or its holiday-worthiness, but the essence of the number itself.

However, partisans of pi, we shall hold strong! Strong as the trees in the forest!

Photo (c) Kenneth Vincent

Now I ask of you, when a person comes across a circle in the woods, what can one measure about that circle? The circumference, of course, by tying a string or tape measure around the outside of the circle. And the diameter, using a tool like a dial caliper (we use these in the engineering classes I teach!). But one cannot measure the radius without tearing apart the circle and violating its circleness by poking at its innards. Besides, even with the careless circumvention of morals that cutting up a circle would entail, how would one determine that circle’s center to then find the radius? One would simply take a diameter and cut it in twain!–there lies our center.

Thus, pi = C/d is much more natural indeed, based on real-world experience, than tau = C/r. Not the reverse, as Ms. Hart claims!

And so, I issue a call to pi devotees everywhere to use this day to trumpet the glories of pi, as we face its enemies near and far! May your circles always be round!

______________________________

NOTE: It has come to my attention that I may be indulging in a bit of hypocrisy with my tirade toward tau. I guess my real opinion is: the more days celebrating pi, its cousins, and all of mathematics, the better. Avery’s conclusion that all of March should be pi month, or Vi’s appeal to celebrate tau day in June, all sound good to me!

In that vein of mathematical mellow-ness, here are a few more pi links that you might enjoy:

• A National Geographic blog post by someone at the Exploratorium (a San Francisco museum that invented or at least popularized pi day).
• A musical composition based on pi. High school friend Brett and I scooped them on this by nine years, using the computers in the band room to compose a pi-digit tune (that wasn’t all that tuneful). The linked video is a very nice rendition of our song, though! 😉
• A pi puzzle (I’ve enjoyed their puzzles each year past, and look forward to working on this one).

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Filed under math

## The Thrill of the Chase

“We need these experiences — noticing patterns, making conjectures, struggling with them mostly unaided by the teacher, researching appropriate related material, crafting a series of flawed-but-each-one-better-than-the-last arguments for why the conjecture could be true, creating a good logical argument and then tweaking it to be airtight — to be the essence of mathematics learning from a young age.”

Appropriately, after writing that yesterday, I got caught up this afternoon in a challenging math question posed by @CmonMattTHINK: “Are there any triangles w/ integer side lengths and exactly one 60° angle?”

If you want to think more about this question yourself, stop reading. I intend to describe my thought processes and answer here, in the interests of thinking through how to create or encourage those thought processes in students.

#### Mathematical Problem-Solving and Proof

These two processes should not be separated, since they are so intimately connected.

I began actually by stumbling halfway into a conversation of other people. @DrMathochist suggested using the law of cosines to create a Diophantine equation:

$c^2 = a^2 + b^2 - ab.$

Then @k8nowak discovered a triple of sides that worked: (3, 8, 7). Note that $3^2 + 8^2 - 3 \cdot 8 = 9 + 64 - 24 = 49 = 7^2$ and that the resulting triangle does indeed have a 60° angle. And @calcdave used the law of sines to express side $a$ in terms of side $b$ and the angle across from $b$ (with a sign error).

Noticing Patterns. At this point, I embarked on my own quest to solve the problem. I noticed that clearly any integer multiples of (3, 8, 7) — like (6, 16, 14) — will also satisfy our equation (of course they will, since they will be similar to the above-linked triangle and therefore also have an angle of 60°). Thus, there are in fact infinitely many triangles that fit @CmonMattTHINK’s question. Rather than stopping there with this answer to the original question, though, it makes sense to ask if there are any other triangles which are not multiples of (3, 8, 7) — that is, going forth we can limit our search to primitive triples wherein the greatest common divisor of the three sides is 1.

Failed Proofs and Stumbling in Wrong Directions. Next, I am embarrassed to say I spent a while using the law of sines (along the track that @calcdave had suggested) to prove that $3=3.$ I should have recognized (but didn’t) when I set two complicated trigonometric expressions equal that had just come from different sides of the law of sines. Then I tried factoring the Diophantine equation above to get $c^2 = (a + \zeta b)(a + \zeta^2 b)$, where $\zeta = \frac{-1+\sqrt{-3}}{2}$ is a cube root of unity. Which led nowhere. Finally, I desperately tried looking at the continued fractions and rational approximations for some irrational numbers that were close to the ratios 3/8, 3/7, etc. No luck. The continued fractions I sought are shown below, using the software Pari/GP.

More Patterns. At this point, I considered writing a program to do a brute force search for more triples. But more handy at the moment was Microsoft Excel, where I could display a number of combinations of $a,$ $b,$ and $a^2 + b^2 - ab,$ and search visually for any that I saw to be perfect squares. I hoped to notice some patterns that could fuel my further work. I found (5, 8, 7), and in general saw that if $(a, b, c)$ is a triple, then $(b-a, b, c)$ will also be a triple (assuming $b>a$). Also encountered were (7, 15, 13) and (5, 21, 19), together with their complements (8, 15, 13) and (16, 21, 19). A piece of my spreadsheet is seen below.

Research. At this point, remembering back to the proof and generation of infinitely many primitive Pythagorean triples from the unit circle, I divided through the Diophantine equation by $c^2$ to get a planar curve $1 = x^2 + y^2 - x y$ on which to look for rational points $(x, y) = (a/c, b/c).$ I graphed this curve in GeoGebra, and plotted the points associated with the triples I had found, as seen here.

I realized that the only rational points on this curve (ellipse?) that I need be concerned with were in the 45° half-quadrant where you see the cluster of points, between (0,1) and (1,1). Clearly negative rational numbers are irrelevant to the sides of a triangle, and the points to the right of (1,1) on the downslope to (1,0) were just rearrangements of which side of the triangle got called $a$ or $b.$

Toward a Solution. OK, we’re in the home stretch now. I pretty much just followed along with what I remembered of the Pythagorean-circle proof. I picked the simplest rational point that I knew was on the curve: (0,1), which doesn’t correspond to any actual triangle, but will soon produce triangles by the method of drawing lines through it. We shall see shortly that lines with rational slope through (0,1) will intersect the curve at rational points (i.e. points where both $x$ and $y$ are rational, and thus yielding solutions to our goals of integer solutions of $c^2 = a^2 + b^2 - a b$ and triangles with angles of 60°).

Note that a slope of 0 through (1,0) would go through (1,1), which represents the equilateral triangle where all angles are 60°. (1,1) is also the point of reflection symmetry, past which we need not go unless we care about the order of $(a,b).$ At (0,1) the curve has instantaneous slope $\frac{dy}{dx} = \frac{1}{2},$ so larger slopes than $\frac{1}{2}$ will intersect the curve in other quadrants. Thus, choosing rational slope $m = p/q \in (0, \frac{1}{2}),$ the line’s equation using good-old point-slope form will be $y = m x + 1.$

Combining this with the curve’s equation and solving for $x,$ we get:

$1 = x^2 + y^2 - x y$

$1 = x^2 + (m x + 1)^2 - x (m x + 1)$

$1 = x^2 + m^2 x^2 + 2 m x + 1 - m x^2 - x$

$0 = x (x + m^2 x + 2 m - m x -1)$

$0 = x + m^2 x + 2 m - m x -1$

$1 - 2m = m^2 x - m x + x$

$x = \frac{1 - 2 m}{m^2 - m + 1}$

Note that $x$ is nonzero because our triangle must have sides that are nonzero.

Although $m$ is rational and constrained to the interval $(0,\frac{1}{2}),$ we’d actually like to express it in terms of natural numbers $p$ and $q,$ where we require that $gcd(p,q)=1$ and $2p < q.$ Leaving aside the messy algebra, we substitute $p/q$ in for $m$ above, and get that:

$x = \frac{q^2 - 2 p q}{p^2 - p q + q^2},$ and $y = \frac{q^2 - p^2}{p^2 - p q + q^2},$ which correspond to sides of the triangle

$(a, b, c) = (q^2 - 2 p q, q^2 - p^2, p^2 - p q + q^2).$

I verified that these expressions do in fact satisfy the relation $c^2 = a^2 + b^2 - a b,$ thus completing the proof that there are infinitely many primitive triples of sides that form a triangle with one 60° angle. [A few easily verifiable details have been left out here.] In fact, not only do we know there are infinitely many, but we can generate them at will. Just pick natural numbers $p$ and $q$ that have no common divisors and where $q$ is more than twice as big as $p.$ Then the triple given above, $(a, b, c) = (q^2 - 2 p q, q^2 - p^2, p^2 - p q + q^2),$ will form a triangle with a single angle of 60°.

The Role of Counterexample in Proof. In trying to prove that this method would always generate a primitive triple (which I believed at first), I actually found a counterexample! For example, $p = 1, q = 5$ yields $(a, b, c) = (15, 24, 21),$ which is a multiple of 3 times the primitive triple (5, 8, 7). But it doesn’t seem that (5, 8, 7) can ever be generated using this method.

I’m pretty sure I have proved that this can only arise if the sum $p + q$ is divisible by 3. If so, this method will generate a triple of numbers thrice a primitive triple. That is, the numbers generated do indeed form a triangle with integer sides and a 60° angle, but which triangle dilated by 1/3 will still satisfy the same criteria. We can call these triple triples, since the triple generated is triple (i.e. three times) a primitive triple.

If the sum $p + q$ is congruent to 1 or 2, modulo 3, then the triple generated will indeed be primitive.

Open Questions (Unknown by me, anyway).

• Is there an alternative method which always generates primitive triples?
• Can my method ever generate (5, 8, 7) or any other of the primitive triples for which a triple triple is generated?

Filed under math, teaching

## Newest Math Teachers at Play!

Hi all!

I expect many readers of this blog may also be familiar with the two math carnivals:  rambling celebrations of all things mathematical that wander monthly among math and math-teacher blogs.  The Carnival of Mathematics (CoM) is explained here; it occurs the first Friday of each month.  Math Teachers at Play (MTaP), with the same period but nicely out-of-phase with the CoM (the cosine to CoM’s sine) occurs on third Fridays of each month, and is housed here.

Well, this months MTaP is up over at Denise’s blog Let’s Play Math.

Here are a few interesting new posts I discovered there:

• Whit poses an interesting answer to the question What is Algebra? “Students should think of algebra not as a set of rules for solving things. . .”
• Meanwhile, Pat presents A Guest Blog (Rant?) from Dave Renfro about Pi Day in the news.
• Eldhose posts a nice assortment of puzzles, too. Kitten impressed me by getting the Coffee machine right, and who wouldn’t chuckle at a rate puzzle which begins: “If a boy and a half eat a hot dog and a half in a minute and a half. . .”
• Now, there are also a ton of other gems up there.  Some I came across before the carnival, for example because they were in my rss reader, or because I follow on Twitter the folks who wrote them.  So please don’t just take my suggestions/discoveries but head on over to Math Teachers at Play to explore on your own!

Lastly, I couldn’t agree more with this recommendation for a fantastic series of New York Times articles on popular mathematics:

• If you haven’t been following the Opinionator math columns, take a look at the latest: Square Dancing. And then go back and read them all.

Filed under math, teaching

When teaching, I try (with mixed results) to keep my students interested in the topic at hand.  Sometimes I motivate them by giving them a project where they discover a mathematical relationship with minimal help and minimal direct instruction from me.  Or a project that demonstrates a deep connection between the real world and the mathematics we use to describe it!

However, sometimes the topic is not all that inherently interesting, or students just need a great deal of practice to really hone a skill.  For example, the skill of solving quadratic equations.  At this point in the course, I’ve already taught them how to solve quadratic equations by factoring, by working backwards, and by the quadratic formula.  To liven things up a bit, I’ll next challenge my students to complete this quadratic equations puzzle (partly pictured at right).

The way it works:

1. Cut up the triangles (in the picture and linked document, they are already mixed, not in their correct final locations).
2. Give each student (or pair of students) a batch of puzzle pieces.
3. Students match each quadratic equation edge to another edge that contains its zeros.  They can tape them together as they go.
4. When complete, it makes a large hexagon, like this:

For extra fun, do the puzzle yourself first, then write a secret message on the backs of the pieces.  Cut it back up into its component pieces, and make double-sided copies so each tile has a letter on its back.  Only students who complete the puzzle correctly will be able to read your secret message!

For example, I have written [one letter to a puzzle piece]:  “BRINGTHISTOMRYFORACOOKIE” (Bring this to Mr. Y. for a cookie!), and then I had cookies available for the students as they finished.  When the first student finishes, brings it up and gets a cookie, the other students sit up and start working harder!

Here’s another puzzle I designed as review of Algebra1-style equations (for my engineering students, to gauge their math strengths and weaknesses in the first week of class).  It could also be used to wrap up an Algebra 1 course, or to refresh students’ memories during the first weeks of Algebra 2.

These puzzles are made with Formulator Tarsia software, which is available for free download.  Developing these puzzles is a bit labor-intensive, in that it takes a good 1.5 – 3 hours to create a good puzzle, including the secret message.

Puzzles (and food) create a nice change of pace in the classroom, every now and then.  I encourage you, gentle reader, to use either of the two I have created, and to download the software to create more yourself!  Happy puzzling!