Monthly Archives: July 2011

Area 10 Squares

Kudos to Dave Gale and chris maths for their great posts about introductory lessons that inspired the questions I pose below. At this point, I don’t have an answer to the query, but I welcome any insights and particularly any other spin-off ideas you may have.

If you have a standard sheet of square grid paper whose dots are exactly 1 unit apart and I ask you to draw a square of area 1, a square of area 4, and a square of area 9, you would probably quickly respond with the following.

Then I ask you to draw a square of area … [deliberate pause] … many immediately begin to think of continuing the pattern to area 16, but instead I ask for a square of area 10. Whether you know the Pythagorean Theorem or just how to compute the areas of squares and triangles, some experimentation hopefully will lead you to some form of the following figure which shows a square with area 10. The real twist for students here is that they need to adjust their point of view from what I’ll call horizontal squares (above) to tilted squares (below).

So, here are my questions:
1) What square areas can be created using square grid paper?
2) What areas of squares can be created more than one way?
3) Is there a largest square area that can NOT be created using square grid paper?

STOP! STOP! STOP!
Do not read any further if you want to work on these questions yourself.

What follows are my musings on these questions and some definite spoilers are included.

Question 1:
Perfect squares (1, 4, 9, 16, 25, 36, …) obviously can be found using increasingly larger horizontal squares. Dave’s ‘blog post gives a great start at the non-perfect, tilted squares.

Image source here.
The information he gives in the image above leads to areas of 2, 5, 8, 10, 13, 17, 18, 20, 25, 26, 29, 32, 34, 37, …

Merging the “tilted” list with the “horizontal list gives 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, 34, 36, 37, …

The missing square areas are 3, 6, 7, 11, 12, 14, 15, 19, 21, 22, 23, 24, …
Whether this list ends depends on the solution to posed question #3.

Question 2:

Comparing the horizontal and tilted lists, the first area that can be found both ways is 25. I briefly thought that was an amazing find until I remembered that the smallest integral Pythagorean Triple is 3-4-5. So a tilted square whose vector position (using Dave’s language) can be expressed as [3,4] can also be created using a horizontal side of 5–the Pythagorean Theorem arises!

There may be other ways to get equivalent square areas (I’d love to hear any if you know some!), but any integral Pythagorean Triple represents a square area that can be represented at least two ways on square grid paper. There are an infinite number of such equivalences.

Question 3:
I don’t know the answer to this, but I think I’m close. I’ll post the problem before I finish it for the fun of letting others into the enjoyment of solving what I think is a cool pre-collegiate level math problem.

I did notice that the missing areas at the end of my discussion of question 1 seems to include a large number of multiples of 3, excluding of course, the horizontal squares with sides that have lengths that are multiples of 3. So is it possible to prove that any area that is
– a multiple of 3, but
– not a perfect square
can not be drawn on square grids?
If so, then there is no maximum area of a square that cannot be drawn using square grid paper.
If not, then the solution to this question may lie in a direction I have not conceived.

Again, I don’t know the answers to questions 1 or 3. Discussion is welcome and encouraged.

Inverses of curves

A fellow Atlanta-area teacher and TI technology buff, Dennis Wilson, started talking to me several months ago about the power of using a locus to describe mathematical relationships. My own exposure to loci had been admittedly limited until that point. When I heard “locus”, my mind immediately went to geometry and the conic sections which I had “learned” by their locus definitions, but I really don’t think I had understood fully how useful a locus was.

The example I give below shows how to construct the inverse to any function using TI-Nspire software. The presentation uses computer software, but easily can be replicated on handheld Nspires. What I particularly like about this approach is that it gives a very strong visual representation of what had largely been a purely algebraic maneuver –switch x & y, and solve for y (if possible). I had always understood inverses as transformations in which the input and output variables reversed roles and knew that was equivalent to reflecting a given curve over y=x, but this approach significantly enhanced my understanding of locus and definitely improved my visual conception of the construction of an inverse.

What I’ve found most amazing is that this approach handles inverses of any curve you can graph in the function menu of the Nspire.

Finding graphs of inverses of non-functions has always been a challenging problem for computers. For example, to find the graph of the inverse of the ellipse using a function graph on an Nspire CAS, you could use the zeros command. Typically, this provides the entire graph, but this inverse approach results in only half of the ellipse and the corresponding half of its inverse. Because the zeros command has limitations for higher order polynomials, I hoped the approach would work for parametrically defined curves. The image below shows that the Geometry Trace still gives the outline of the curve (probably enough for most), but the locus doesn’t work. (I suspect this is because I’m creating the inverse from _dependent_ x- and y-coordinates.)

The same restrictions (probably for the same dependence-independence reasons) apply to polar curves, but at least you can trace the inverse.

I need to think more, but for now, thank you, Dennis, for opening a door in my mind that should have been opened long ago.

Birthday Problem redux

A former student (Elizabeth) sent me a note.  She asked,

“I was looking at a limited lineage of my family tree of my ancestors…. Out of 18 individuals and 365 days out of the year, only one date was repeated as a day these individuals were born, and it was repeated 3 times. The day was February 13. Can you calculate the probability of this event?”

There are MANY good math problems here.  I’ll start with Elizabeth’s direct question and offer some alternatives.  All of my answers use 365 days/year (ignoring leap years) and assume all birthdays are equally likely.  Both of these have some problems, but it makes the solutions a bit easier.

  1. What is the probability of exactly 3 people out of 18 being born on February 13 and no others sharing a birthday? 
    Of the 18 relatives, pick the 3 who share the February 13 birthday.  This can happen in \displaystyle C\left(18,3\right)=\frac{18!}{(18-3)!\cdot 3!}=816 ways. The other 15 relatives have 15 different birthdays in the remaining 364 days of the year.  This can happen in \displaystyle P\left(364,15\right)\approx 1.95\cdot 10^{38} ways (This is a big number!).  So the probability of Elizabeth’s exact query is \displaystyle \frac{C\left(18,3\right) \cdot P\left(364,15\right)}{365^{18}}\approx 1.20\cdot 10^{-5} , or a little more than 1 chance in 83,333.
  2. What is the probability of exactly 3 people out of 18 being born on the same day and no others sharing a birthday?
    The difference now is that the shared birthday could be anywhere in the calendar.  There are 365 ways that could happen with the rest of the problem being the same. This new probability is \displaystyle \frac{365\cdot C\left(18,3\right) \cdot P\left(364,15\right)}{365^{18}}\approx 0.00439 , roughly 1 chance in 278.
  3. What is the probability that nobody in a group of 18 people shares a birthday?
    That means choosing 18 of the 365 birthdays.  This probability is \displaystyle \frac{P\left(365,18\right)}{365^{18}}\approx 0.653=65.3\%.

Perhaps much more interesting than the last question (to me anyway) is that this means there is a 1-0.653=34.7\% chance that there are shared birthdays in any group of 18 people–a surprisingly high percentage to most people.  This is closely related to the “famous” birthday problem for which there are MANY online explanations (e.g., here and here).

QUICK MATH TRIVIA:  In any random group of 22 or more people, there is at least a 50% chance that two people will share the same birthday.  Thanks to knowledge of the birthday problem, this could win you some surprisingly easy bets!

If you want to read more about the probabilities of triplets of birthdays, I suggest this Math Forum post.  Make sure you read the last entry by Doctor Rick; the first answer is not quite correct.

Wolfram Alpha’s discussion of the Birthday Problem moves very quickly to a generalized solution if you can stomach lots of symbols.

The complexity of the computations mentioned in the WA article are necessary if you consider the number of ways birthday pairings could occur.  In all cases, birthday problem solutions almost always boil down to computing the number of ways for an event to NOT occur and subtracting the result from 1.

Combinatorics problems like these are deceptively simple to state in words and are often notoriously complicated to state mathematically.  Once you have a good framing of the problem, though, they can be straightforward to compute, even if cumbersome.

BIAS NOTE:  All of these computations assume all birthdays are equally likely.  This simply isn’t true. Roy Murphy’s analysis (here) concludes the following.

The months July – October show higher than expected births and March – May show the most significant decline in births. Perhaps the most reasonable explanation is that conceptions are up in the months of October through January and down in June through August. You be the judge.

I thought this infographic was an especially cool demonstration:

Great mathematics problems just keep on giving.