Tag Archives: algebra

Powers of i

I was discussing integer powers of i in my summer Algebra 2 last month and started with the “standard” modulus-4 pattern I learned as a student and have always taught.  While not particularly insightful, my students and I considered another approach that might prove simpler for some.


I began with the obvious i^0 and i^1 before invoking the definition of i to get i^2.  From these three you can see every time the power of i increases by 1, you multiply the result by i and simplify the result if possible using these first 3 terms.  The result of i^3 is simple,  taking the known results to


But i^4=-i^2=-(-1)=1, cycling back to the value initially found with i^0.  Continuing this procedure creates a modulus-4 pattern:


They noticed that to any multiple of 4 was 1, and other powers were i, -1, or –i, depending on how far removed they were from a multiple of 4.  For an algorithm to compute a simplified form of to an integer power, divide the power by 4, and raise i to the remainder (0, 1, 2, or 3) from that division.

They got the pattern and were ready to move on when one student who had glimpsed this in a math competition at some point noted he could “do it”, but it seemed to him that memorizing the list of 4 base powers was a necessary requirement to invoking the pattern.

Then recalled a comment I made on the first day of class.  I value memorizing as little mathematics as possible and using the mathematics we do know as widely as possible.  His challenge was clear:  Wasn’t asking students to use this 4-cycle approach just a memorization task in disguise?  If I believed in my non-memorization claim, shouldn’t there be another way to achieve our results using nothing more the definition of i?


By definition, i = \sqrt{-1}, so it’s a very small logical stretch with inverse operations to claim i^2=-1.

Even Powers:  After trying some different examples, one student had an easy way to handle even powers.  For example, if n=148, she invoked an exponent rule “in reverse” to extract an i^2 term which she turned into a -1.  Because -1 to any integer power is either 1 or -1, she used the properties of negative numbers to odd and even powers to determine the sign of her answer.


Because any even power can always be written as the product of 2 and another number, this gave an easy way to handle half of all cases using nothing more than the definition of i and exponents of -1.

A third student pointed out another efficiency.  Because the final result depended only on whether the integer multiplied by 2 was even or odd, only the last two digits of n were even relevant.  That pattern also exists in the 4-cycle approach, but it felt more natural here.

Odd Powers:  Even powers were so simple, they were initially frustrated that odd powers didn’t seem to be, too.  Then the student who’d issued the memorization challenge said that any odd power of i was just the product of i and an even power of i.  Invoking the efficiency in the last paragraph for n=567, he found



In the end, powers of i had become nothing more complicated than exponent properties and powers of -1.  The students seemed to have greater comfort with finding powers of complex numbers, but I have begun to question why algebra courses have placed so much emphasis on powers of i.

From one perspective, a surprising property of complex numbers for many students is that any operation on complex numbers creates another complex number.  While they are told that complex numbers are a closed set, to see complex numbers simplify so conveniently surprises many.

Another cool aspect of complex number operations is the stretch-and-rotate graphical property of complex number multiplication.   This is the basis of DeMoivre’s Theorem and explains why there are exactly 4 results when you repeatedly multiply any complex number by i–equivalent to stretching by a factor of 1 and rotating \frac{\pi}{2}.  Multiplying by 1 doesn’t change the magnitude of a number, and after 4 rotations of \frac{\pi}{2}, you are back at the original number.

So, depending on the future goals or needs of your students, there is certainly a reason to explore the 4-cycle nature of repeated multiplication by i.  If the point is just to compute a result, perhaps the 4-cycle approach is unnecessarily “complex”, and the odd/even powers of -1 is less computationally intense.  In the end, maybe it’s all about number sense.

My students discovered a more basic algorithm, but I’m more uncomfortable.  Just because we can ask our students a question doesn’t mean we should.  I can see connections from my longer studies, but do they see or care?  In this case, should they?

From Coins to Magic

Here’s a great problem or trick for a class exploration … or magic for parties.


Grab a small handful of coins (it doesn’t matter how many), randomly flip them onto a flat surface, and count the number of tails.

Randomly pull off from the group into a separate pile the number of coins equal to the number of tails you just counted.  Turn over every coin in this new pile.

Count the number of tails in each pile.

You got the same number both times!



Marilyn Vos Savant posed a similar problem:

Say that a hundred pennies are on a table. Ninety of them are heads. With your eyes closed, can you separate all the coins into two groups so that each group has the same number of tails?

Savant’s solution is to pull any random 10 coins from the 100 and make a second pile.  Turn all the coins in the new pile over, et voila!  Both piles have an equal number of tails.

While Savant’s approach is much more prescriptive than mine, both solutions work.  Every time.  WHY?


You have no idea the state (heads or tails) of any of the coins you pull into the second pile.  It’s counterintuitive that the two piles could ever contain the same number of tails.

Also, flipping the coins in the new pile seems completely arbitrary, and yet after any random pull & flip, the two resulting piles always hold the same number of tails.

Enter the power (and for young people, the mystery) of algebra to generalize a problem, supporting an argument that holds for all possibilities simultaneously.


The first clue to this is the misdirection in Savant’s question.  Told that there are 90 heads, you are asked to make the number of tails equivalent.  In both versions, the number of TAILS in the original pile is the number of coins pulled into the second pile.  This isn’t a coincidence; it’s the key to the solution.

In any pile of randomly flipped coins (they needn’t be all or even part pennies), let N be the number tails.  Create your second pile by pulling a random coins from the initial pile.  Because the coins are randomly selected, you don’t know how many tails are in the new pile, so let that unknown number of coins be X .  That means 0 \le X \le N, leaving N-X tails in the first pile, and N-X heads in the new pile.  (Make sure you understand that last bit!)  That means if you flip all the coins in the second pile, those heads will become tails, and you are guaranteed exactly N-X tails in both piles.

Cool facts:

  • You can’t say with certainty how many tails will be in both piles, but you know they will be the same.
  • The total number of coins you start with is completely irrelevant.
  • While the given two versions of the problem make piles with equal numbers of heads, this “trick” can balance heads or tails.  To balance heads instead, pull from the initial coins into a second pile the number of heads.  When you flip all the coins in the second pile, both piles will now contain the same number of heads.


If you work on your showmanship, you can baffle others with this.  For my middle school daughter, I counted off the “leave alone” pile and then flipped the second pile.  I also let her flip the initial set of coins and tell me each time whether she wanted me to get equal numbers of heads or tails.  I looked away as she shuffled the coins and pulled off the requisite number of coins without looking.

She’s figured out HOW I do it, but as she is just starting algebra, she doesn’t have the abstractness yet to fully generalize the big solution.  She’ll get there.

I could see this becoming a fun data-gathering project for an algebra class.  It would be cool to see how someone approaches this with a group of students.


Party Ratios

I find LOTS of great middle school problems from @Five_Triangles on Twitter.  Their post two days ago was no exception.

The problem requires a little stamina, but can be approached many ways–two excellent criteria for worthy student explorations.  That it has some solid extensions makes it even better.  Following are a few different solution approaches some colleagues and I created.


The most challenging part of this problem is data organization.  My first thoughts were for a 2-circle Venn Diagram–one for gender and one for age.  And these types of Venn Diagrams are often more easily understood, in my experience, in 2×2 Table form with extra spaces for totals.  Here’s what I set up initially.


The ratio of Women:Girls was 11:4, so the 24 girls meant each “unit” in this ratio accounted for 24/4=6 people.  That gave 11*6=66 women and 66+24=90 females.


At this point, my experience working with algebraic problems tempted me to overthink the situation.  I was tempted to let B represent the unknown number of boys and set up some equations to solve.  Knowing that most 6th graders would not think about variables, I held back that instinct in an attempt to discover what a less-experienced mind might try. I present my initial algebra solution below.

The 5:3 Male:Female ratio told me that each “gender unit” represented 90/3=30 people.  That meant there were 5*30=150 men and 240 total people at the party.


Then, the 4:1 Adult:Children ratio showed how to age-divide every group of 5 partygoers.  With 240/5=48 such groups, there were 48 children and 4*48=192 adults.  Subtracting the already known 66 women gave the requested answer:  192-66=126 men.


While this Venn Diagram/Table approach made sense to me, I was concerned that it was a moderately sophisticated and not quite intuitive problem-solving technique for younger middle school students.


A middle school teaching colleague, Becky, offered a different solution I could see students creating.

Completely independently, she solved the problem in exactly the same order I did using ratio tables to manage the scaling at each step instead of my “unit ratios”.  I liked her visual representation of the 4:1 Adults:Children ratio to find the number of adults, which gave the requested number of men.  I suspect many more students would implicitly or explicitly use some chunking strategies like the visual representation to work the ratios.




Math problems involving ratios can usually be opened up to allow multiple, or even an infinite number of solutions.  This leads to some interesting problem extensions if you eliminate the “24 girls” restriction.  Here are a few examples and sample solutions.

What is the least number of partygoers?

For this problem, notice from the table above that all of the values have a common factor of 6.  Dividing the total partygoers by this reveals that 240/6=40 is the least number.  Any multiple of this number is also a legitimate solution.

Interestingly, the 11:4 Women:Girls ratio becomes explicitly obvious when you scale the table down to its least common value.

ratio6My former student and now colleague, Teddy, arrived at this value another way.  Paraphrasing, he noted that the 5:3 Male:Female ratio meant any valid total had to be a multiple of 5+3=8.  Likewise, the 4:1 Adult:Child ratio requires totals to be multiples of 4+1=5.  And the LCM of 8 & 5 is 40, the same value found in the preceding paragraph.

What do all total partygoer numbers have in common?

As explained above, any multiple of 40 is a legitimate number of partygoers.

If the venue could support no more than 500 attendees, what is the maximum number of women attending?

12*40=480 is the greatest multiple of 40 below 500.  Because 480 is double the initial problem’s total, 66*2=132 is the maximum number of women.

Note that this can be rephrased to accommodate any other gender/age/total target.

Under the given conditions, will the number of boys and girls at the party ever be identical?

As with all ratio problems, larger values are always multiples of the least common solution.  That means the number of boys and girls will always be identical or always be different.  From above, you can deduce that the numbers of boys and girls at the party under the given conditions will both be multiples of 4.

What variations can you and/or your students create?


Now to the solution variation I was initially inclined to produce.  After initially determining 66 women from the given 24 girls, let B be the unknown number of boys.  That gives B+24 children.  It was given that adults are 4 times as numerous as children making the number of adults 4(B+24)=4B+96.  Subtracting the known 66 women leaves 4B+30 men.  Compiling all of this gives


The 5:3 Male:Female ratio means \displaystyle \frac{5}{3} = \frac{5B+30}{90} \longrightarrow B=24, the same result as earlier.



Winding through all of that algebra ultimately isn’t that computationally difficult, but it certainly is more than typical 6th graders could handle.

But the problem could be generalized even further, as Teddy shared with me.  If the entire table were written in variables with W=number of women, M=men, G=girls, and B=boys, the given ratios in the problem would lead to a reasonably straightforward 4×4 system of equations.  If you understand enough to write all of those equations, I’m certain you could solve them, so I’d feel confident allowing a CAS to do that for me.  My TI-Nspire gives this.


And that certainly isn’t work you’d expect from any 6th grader.


Given that the 11:4 Women:Girls ratio was the only “internal” ratio, it was apparent in retrospect that all solutions except the 4×4 system approach had to find the female values first.  There are still several ways to resolve the problem, but I found it interesting that while there was no “direct route”, every reasonable solution started with the same steps.

Thanks to colleagues Teddy S & Becky M for sharing their solution proposals.

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.


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?


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.


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.


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.


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!


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.

Chemistry, CAS, and Balancing Equations

Here’ s a cool application of linear equations I first encountered about 20 years ago working with chemistry colleague Penney Sconzo at my former school in Atlanta, GA.  Many students struggle early in their first chemistry classes with balancing equations.  Thinking about these as generalized systems of linear equations gives a universal approach to balancing chemical equations, including ionic equations.

This idea makes a brilliant connection if you teach algebra 2 students concurrently enrolled in chemistry, or vice versa.


Consider burning ethanol.  The chemical combination of ethanol and oxygen, creating carbon dioxide and water:

C_2H_6O+3O_2 \longrightarrow 2CO_2+3H_2O     (1)

But what if you didn’t know that 1 molecule of ethanol combined with 3 molecules of oxygen gas to create 2 molecules of carbon dioxide and 3 molecules of water?  This specific set coefficients (or multiples of the set) exist for this reaction because of the Law of Conservation of Matter.  While elements may rearrange in a chemical reaction, they do not become something else.  So how do you determine the unknown coefficients of a generic chemical reaction?

Using the ethanol example, assume you started with

wC_2H_6O+xO_2 \longrightarrow yCO_2+zH_2O     (2)

for some unknown values of w, x, y, and z.  Conservation of Matter guarantees that the amount of carbon, hydrogen, and oxygen are the same before and after the reaction.  Tallying the amount of each element on each side of the equation gives three linear equations:

Carbon:  2w=y
Hydrogen:  6w=2z
Oxygen:  w+2x=2y+z

where the coefficients come from the subscripts within the compound notations.  As one example, the carbon subscript in ethanol ( C_2H_6O ) is 2, indicating two carbon atoms in each ethanol molecule.  There must have been 2w carbon atoms in the w ethanol molecules.

This system of 3 equations in 4 variables won’t have a unique solution, but let’s see what my Nspire CAS says.  (NOTE:  On the TI-Nspire, you can solve for any one of the four variables.  Because the presence of more variables than equations makes the solution non-unique, some results may appear cleaner than others.  For me, w was more complicated than z, so I chose to use the z solution.)


All three equations have y in the numerator and denominators of 2.  The presence of the y indicates the expected non-unique solution.  But it also gives me the freedom to select any convenient value of y I want to use.  I’ll pick y=2 to simplify the fractions.  Plugging in gives me values for the other coefficients.


Substituting these into (2) above gives the original equation (1).


Traditionally, chemists write these equations with the lowest possible natural number coefficients, but thinking of them as systems of linear equations makes another reality obvious.  If 1 molecule of ethanol combines with 3 molecules of hydrogen gas to make 2 molecules of carbon dioxide and 3 molecules of water, surely 10 molecule of ethanol combines with 30 molecules of hydrogen gas to make 20 molecules of carbon dioxide and 30 molecules of water (the result of substituting y=20 instead of the y=2 used above).

You could even let y=1 to get z=\frac{3}{2}, w=\frac{1}{2}, and x=\frac{3}{2}.  Shifting units, this could mean a half-mole of ethanol and 1.5 moles of hydrogen make a mole of carbon dioxide and 1.5 moles of water.  The point is, the ratios are constant.  A good lesson.


Now let’s try a harder one to balance:  Reacting carbon monoxide and hydrogen gas to create octane and water.

wCO + xH_2 \longrightarrow y C_8 H_{18} + z H_2 O

Setting up equations for each element gives

Carbon:  w=8y
Oxygen:  w=z
Hydrogen:  2x=18y+2z

I could simplify the hydrogen equation, but that’s not required.  Solving this system of equations gives


Nice.  No fractions this time.  Using y=1 gives w=8, x=17, and z=8, or

8CO + 17H_2 \longrightarrow C_8 H_{18} + 8H_2 O



Now let’s balance an ionic equation with unknown coefficients a, b, c, d, e, and f:

a Ba^{2+} + b OH^- + c H^- + d PO_4^{3-} \longrightarrow eH_2O + fBa_3(PO_4)_2

In addition to writing equations for barium, oxygen, hydrogen, and phosphorus, Conservation of Charge allows me to write one more equation to reflect the balancing of charge in the reaction.

Barium:  a = 3f
Oxygen:  b +4d = e+8f
Hydrogen:  b+c=2e
Phosphorus:  d=2f
CHARGE (+/-):  2a-b-c-3d=0

Solving the system gives


Now that’s a curious result.  I’ll deal with the zeros in a moment.  Letting d=2 gives f=1 and a=3, indicating that 3 molecules of ionic barium combine with 2 molecules of ionic phosphate to create a single uncharged molecule of barium phosphate precipitate.

The zeros here indicate the presence of “spectator ions”.  Basically, the hydroxide and hydrogen ions on the left are in equal measure to the liquid water molecule on the right.  Since they are in equal measure, one solution is

3Ba^{2+}+6OH^- +6H^-+2PO_4^{3-} \longrightarrow 6H_2O + Ba_3(PO_4)_2


You still need to understand chemistry and algebra to interpret the results, but combining algebra (and especially a CAS) makes it much easier to balance chemical equations and ionic chemical equations, particularly those with non-trivial solutions not easily found by inspection.

The minor connection between science (chemistry) and math (algebra) is nice.

As many others have noted, CAS enables you to keep your mind on the problem while avoiding getting lost in the algebra.

Circle and Square

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

Following is the problem as Justin posed it yesterday.


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


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


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

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


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


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

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

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

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

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


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

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


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


Powers of 2

Yesterday, James Tanton posted a fun little problem on Twitter:

2powersSo, 2 is one more than 1=2^0, and 8 is one less than 9=2^3$, and Dr. Tanton wants to know if there are any other powers of two that are within one unit of a perfect square.

While this problem may not have any “real-life relevance”, it demonstrates what I describe as the power and creativity of mathematics.  Among the infinite number of powers of two, how can someone know for certain if any others are or are not within one unit of a perfect square?  No one will ever be able to see every number in the list of powers of two, but variables and mathematics give you the tools to deal with all possibilities at once.

For this problem, let D and N be positive integers.  Translated into mathematical language, Dr. Tanton’s problem is equivalent to asking if there are values of D and N for which 2^D=N^2 \pm 1.  With a single equation in two unknowns, this is where observation and creativity come into play.  I suspect there may be more than one way to approach this, but my solution follows.  Don’t read any further if you want to solve this for yourself.


Because D and N are positive integers, the left side of 2^D=N^2 \pm 1,  is always even.   That means N^2, and therefore N must be odd.

Because N is odd, I know N=2k+1 for some whole number k.  Rewriting our equation gives 2^D=(2k+1)^2 \pm 1, and the right side equals either 4k^2+4k or 4k^2+4k+2.

Factoring the first expression gives 2^D=4k^2+4K=4k(k+1).   Notice that this is the product of two consecutive integers, k and k+1, and therefore one of these factors (even though I don’t know which one) must be an odd number.  The only odd number that is a factor of a power of two is 1, so either k=1 or k+1=1 \rightarrow k=0.  Now, k=1 \longrightarrow N=3 \longrightarrow D=3 and k=0 \longrightarrow N=1 \longrightarrow D=0, the two solutions Dr. Tanton gave.  No other possibilities are possible from this expression, no matter how far down the list of powers of two you want to go.

But what about the other expression?  Factoring again gives 2^D=4k^2+4k+2=2 \cdot \left( 2k^2+2k+1 \right) .  The expression in parentheses must be odd because its first two terms are both multiplied by 2 (making them even) and then one is added (making the overall sum odd).  Again, 1 is the only odd factor of a power of two, and this happens in this case only when k=0 \longrightarrow N=1 \longrightarrow D=0, repeating a solution from above.

Because no other algebraic solutions are possible, the two solutions Dr. Tanton gave in the problem statement are the only two times in the entire universe of perfect squares and powers of two where elements of those two lists are within a single unit of each other.

Math is sweet.