## Posts tagged ‘square’

### What an Amazing Date!

There are lots of good dates — meeting at a bookstore coffee shop for, well, perusing books and sipping coffee; spending hours playing Super Mario Bros. and Pac-Man at a retro video game arcade; and, of course, going to an open-mic comedy show where one of the performers tells nothing but math jokes.

But great dates? Well, those are pretty rare. My first date with my wife — where I took her to a hotel and “whispered” to her from the couch across the lobby — is an example, though the stimulating conversation and her perfect laugh may have contributed more than the elliptical ceiling. (Maybe.)

Few dates, however, can compare to today’s date:

**12/3/21**

Look at that beautiful symmetry! Marvel at its palindromic magnificence! The way it rises then falls, like a Shostakovich melody.

But wait… there’s more! Consider the following pattern:

1 × 1 = 1

11 × 11 = 121

111 × 111 = ?

That’s right! The number 12,321 is a perfect square! And not only that, its square root contains only 1s.

Moreover, check this out:

1 + 2 + 3 + 2 + 1 = 9

That’s right! It’s a square number, and the sum of its digits is also a square number!

Finally, here’s a KenKen puzzle that makes use of the number, though it’s not unique unless one of the digits is already filled in:

No matter how you choose to celebrate, here’s hoping your day is as great as the date!

### Covering 100 Squares

There’s an old math joke that says math books are sad because they have too many problems. But I disagree; I believe that most math books — and, in particular, textbooks — are sad because they have too many *exercises*.

To try to release more true problems into the wild, I recently wrote a book called *One-Hundred Problems Involving the Number 100*. Published by NCTM, the book contains problems, suggestions for classroom use, and solutions. Some of the problems are old chestnuts, such as, “What is 1 + 2 + 3 + ⋯ + 100?” (Problem 21: Gauss and Check). Others are complete originals, and occasionally a little silly, such as, “If all the positive integers from 1 to 100 were spelled out, how many letters would be used?” (Problem 1: Spell It Out). But my favorite problem in the book, which I’ve shared twice during NCTM’s 100 Days of Professional Development (**May 14**, **Oct 7**; login required), is Problem 100: Covering with Squares.

**As shown below, a square grid with 100 smaller squares can be covered by 100 squares (each measuring 1 × 1), by 25 squares (each measuring 2 × 2), or by 13 squares (one 6 × 6, two 4 × 4, two 3 × 3, two 2 × 2, and six 1 × 1).**

**Find all values of 0 < n < 100 for which it is impossible to cover a 10 × 10 grid with n squares of integer side length.**

I’ll admit, I’m not ecstatic about the phrasing of the question at the end. The symbolic representation makes it succinct, sure, but this problem can be investigated by students who may be too young to understand that notation. It might be better to just ask, “Can you cover it with two squares? Can you cover it with 58 squares? For how many different numbers of squares can you find a covering?”

You can, of course, explore this problem using graph paper. But for a digital experience, click on the link below and explore using Google Slides:

When you click on that link, you’ll be required to “Make a copy” for your own Google Drive. This is done for two reasons: first, it’ll prohibit any editing to the original file, so that other folks who use that link will receive a similarly pristine copy; and second, it places a copy in your Google Drive that you can play with, modify, or share with your students. If you do that, be sure to change **/edit** and anything that follows at the end of the URL to **/copy**, which will require anyone you share it with to make a copy, too. Of course, you could also just share the link above with your students, and they can have a copy of the original file.

When using this problem in the classroom, start by showing a few examples — such as the ones above — and then have all students find a covering for the grid that’s different from your examples. Because every student can find at least one covering, this breeds confidence. In addition, it ensures that all students understand the problem. When all students have found a covering, have them share it with a partner. From then on, allow students to work in pairs or small groups to find other coverings.

To facilitate the discussion, I draw a blank hundreds grid on the whiteboard. As unique coverings are found, students are allowed to enter the number in the hundreds grid. Of course, I require them to first verify with a partner that they have counted the number of squares correctly; once confirmed, they can show the covering to me; and if that number hasn’t been entered in the hundreds grid yet, they get to enter it. This can be motivating for students, and it’s a good opportunity to get less participatory students engaged.

When using this problem recently with a group of students, this is how the hundreds grid looked after about 20 minutes:

You’ll notice that numbers appear in different colors; each group received a different color dry erase marker. What you might also notice is that the numbers 17, 20, 23, 26, …, 65 all appear in red. This seemed more than coincidental, so I asked about it. The group members explained:

*We realized we could cover the grid with a one 6 × 6 square and sixteen 2 × 2 squares:*

*We then divided one of the 2 × 2 squares into four 1 × 1 squares. That meant that one 2 × 2 square was replaced by four 1 × 1 squares, increasing the number of squares by three. So, the grid was now covered with 17 + 3 = 20 squares. *

*If we kept doing that — if we kept dividing the 2 × 2 squares — then we’d keep adding three more squares, so we could get 23, 26, 29, and so on, all the way up to 65.*

The realization that one configuration could be transformed into many others allowed students to find coverings for myriad numbers. For what values of *n* could the grid not be covered? That’s left as a question for you. Have fun!

### It’s Not What’s on the Outside…

Through the Academic and Creative Endeavors (ACE) program at their school, my sons participate in the Math Olympiad for Elementary and Middle School (MOEMS). While passing the door to the ACE room yesterday, I noticed a sign with the names of those who scored a perfect 5 out of 5 on the most recent contest — and my sons’ names were conspicuously absent. Last night at dinner, I asked Alex and Eli what happened, and they told me about the problem that they both missed. (What? Like we’re the only family in America that discusses math problems at the dinner table.) Here’s how they explained it to me:

Nine 1-cm by 1-cm squares are arranged to form a 3 × 3 square, as shown below. The 3 × 3 square is divided into two pieces by cutting along gridlines only. What is the greatest total (combined) perimeter for the two shapes?

*The answer to this problem appears below. Pause here if you’d like to solve it before reaching the spoiler.*

[Ed. Note: I didn’t see the actual exam, so the presentation of the problem above is based entirely on my sons’ description. Apologies to MOEMS for any substantive differences.]

This problem epitomizes what I love about math competitions.

**The answer to the problem is not obvious.**This is the case with many competition problems, unlike the majority of problems that appear in a traditional textbook.**The solution does not rely on rote mechanics.**Again, this differentiates it from a standard textbook problem or — shudder! — from the problems that often populate the databases of many skill-based online programs.**Students have to get messy.**That is, they’ll need to try something, see what happens, then decide if they can improve the result.**Students have to convince themselves when to stop.**Or more precisely, they’ll need to convince themselves that they’ve found the correct answer. For instance, let’s say a student divides the square into a 3 × 2 and a 3 × 1 rectangle. The combined perimeter is 18 units. Is that good enough, or can you do better? This is different from, say, a typical algebra problem, for which students are taught how to check their answer.

The problem also epitomizes what many people hate about math competitions.

**There’s a time limit.**Students have 26 minutes to solve 5 problems. Which means that if students spend more than 5 minutes on this one, they may not have time to finish the other four. (There was a student of John Benson who, when asked about his goal for an upcoming math competition, replied, “I hope to solve half the problems during the competition and all of them by the end of the week.” That’s the way mathematicians work.)**It’s naked math.**Sorry, nothing real-world about this one. (But maybe that’s okay, because real may not be better.)**The problem is presented as a neat little bundle.**This is rarely how mathematics actually works. True problems often don’t present themselves all at once; it’s through investigation and research that the constraints become known and the nuances are revealed.

All that said, *I believe that the pros far outweigh the cons*. Benjamin Franklin Finkel said, “Many dormant minds have been aroused into activity through the mastery of a single problem.” I don’t remember the last time a mind was aroused by the solution to *x* + 7 = -3, but I’ve witnessed awakenings when students solve problems like the one above.

And here’s the tragedy in all of this: **Many teachers believe that only kids who participate in math competitions can handle — or appreciate — math competition questions.** No! Quite the opposite, in fact. Students who have tuned out have done so because they’ve never been challenged and, worse, have never felt the thrill of solving a problem on their own.

What I really love about the problem, though, is it made me think about other questions that could be asked:

- How many ways are there to divide a 3 × 3 square into two pieces that will yield the maximum total perimeter?
- What is the maximum total perimeter if a 3 × 3 square is divided into three pieces?
- What is the maximum total perimeter if a 4 × 4 square is divided into two pieces? …a 5 × 5 square? …a 6 × 6 square?
*The answer to the problem about the 4 × 4 square appears below. Pause here if you’d like to solve it before reaching the spoiler.*

- (wait for it) What is the maximum total perimeter if an
*n*×*n*square is divided into two pieces?

It was that last question that really got the blood pumping.

Here’s a solution for how to divide a 3 × 3 square to yield the greatest total perimeter:

And here’s a solution for how to divide a 4 × 4 square to yield the greatest total perimeter:

What’s the solution for an *n* × *n* square? That’s left as an exercise for the reader.

### Mathiest Fortnight of 2016

Monday, April 4, 2016, was Square Root Day, because the date is abbreviated 4/4/16, and 4 × 4 = 16. But if you’re a faithful reader of this blog, then you already knew that, because you read all about it in Monday’s post, Guess the Graph on Square Root Day.

But it doesn’t end there. It ain’t just one day. Oh, no, friends… this is a banner week. Or, really, a banner *two weeks*.

Tomorrow, April 8, 2016, is a **geometric sequence day**, because the date is 4/8/16, and 4 × 2 = 8, and 8 × 2 = 16.

And Saturday, April 9, 2016, is a **consecutive square number day**, because the month, day, and year are consecutive square numbers. Square number days, in which each of the month, day, and year are all square numbers — not necessarily consecutive — are less rare; there are 15 of them this year. But among them is 1/4/16, which rocks the intersection of square number days *and* geometric sequence days. (That’s right — I said “rocks the intersection.”)

And then Sunday, April 10, 2016, is an **arithmetic sequence day**, because 4, 10, and 16 have a common difference of 6. Though honestly, arithmetic sequence days are a dime a half-dozen; there are six of them this year.

Next Monday, April 12, 2016, is a **sum day**, because 4 + 12 = 16. Again, ho-hum. There are a dozen sum days this year, and there will be a dozen sum days *every year* through 2031.

And just a little further in the future is Friday, April 16, 2016, whose abbreviation is 4/16/16, and if you remove those unsightly slashes, you get 41,616 = 204^{2}. I’m not sure what you’d call such a day, other than *awesome*.

Admittedly, some of those things are fairly common occurrences. But, still. That’s six calendar-related phenomena in a thirteen-day period, which may be enough mathematic-temporal mayhem to unseat the previously unrivaled Mathiest Week of 2013.

Partially, this blog post was meant to enlighten and entertain you. But mostly, it was meant to send numerologists off the deep end. Mission. Accomplished.

You’ve endured enough. Here are some calendar-related jokes for you…

Did you hear about the two grad students who stole a calendar?

They each got six months!I was going to look for my missing calendar, but I just couldn’t find the time.

What do calendars eat?

Dates.

### A Father’s Day Gift Worth Waiting For

Alex made a Father’s Day Book for me. Because the book didn’t make it on our trip to France, however, I didn’t receive it until this past weekend. It was worth the wait.

The book was laudatory in praising my handling of routine fatherly duties:

I loved when you took me to Smashburger.

I appreciated when you helped me find a worm.

I love when you read to me at night.

I love when I see you at the sign-out sheet [at after-school care]. It means I can spend time with you.

But my favorite accolade — surprise! — was mathematical:

I liked the multiplication trick you taught me. Take two numbers, find the middle [average], square it. Find the difference [from one number to the average], square it, subtract it. (BOOM! Done!)

Priceless.

The trick that I taught him was how to use the difference of squares to quickly find a product. For instance, if you want to multiply 23 × 17, then…

- The average of 23 an 17 is 20, and 20
^{2}= 400. - The difference between 23 and 20 is 3, and 3
^{2}= 9. - Subtract 400 – 9 = 391.
- So, 23 × 17 = 391.
**BOOM! Done!**

This works because

,

and if you let *a* = 20 and *b* = 3, then you have

.

In particular, I suggested this method if (1) the numbers are relatively small and (2) either both are odd or both are even. I would not recommend this method for finding the product 6,433 × 58:

- The average is 3,245.5, and (3,245.5)
^{2}= 10,533,270.25. - The difference between 6,433 and 3,245.5 is 3,187.5, and (3,187.5)
^{2}= 10,160,156.25. - Subtract 10,533,270.25 – 10,160,156.25 = 373,114.
- So, 6,433 × 58 = 373,114.

Sure, it works, but that problem screams for a calculator. The trick only has utility when the numbers are small and nice enough that finding the square of the average and difference is reasonable.

Then again, it’s not atypical for sons to do unreasonable things…

Son: Would you do my homework?

Dad: Sorry, son, it wouldn’t be right.

Son: That’s okay. Can you give it a try, anyway?

I’m just glad that my sons understand math at an abstract level…

A young boy asks his mother for some help with math. “There are four ducks on a pond. Two more ducks join them on the pond. How many ducks are there?”

The mother is surprised. She asks, “You don’t know what 4 + 2 is?”

“Sure, I do,” says the boy. “It’s 6. But what does that have to do with ducks?”

### Celebrity Sighting at Math Meeting

I have the pleasure of serving on the advisory committee for the Math Midway 2 Go, a traveling exhibit of the Museum of Mathematics. I get to see a lot of cool stuff.

When it opens on December 15, MoMath will be the only museum of mathematics in North America. If you happen to find yourself in Manhattan, check it out. The exhibits are really fun.

One of the exhibits in the Math Midway 2 Go is a number line with ornaments hanging from each number. For instance, a square ornament hangs from the numbers 1, 4, 9, 16, …, and a symbol that looks like an atom hangs from 2, 3, 5, 7, 11, 13, … (the atom symbol was used because “prime numbers are the building blocks of the number system”). However, I was not able to identify the symbol that hangs from the following numbers:

3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30,

32, 34, 40, 48, 51, 60, 64, 68, 80, 85, 96

I’ve been told that the symbol is a compass (the kind for drawing circles, not for orienteering). Unfortunately, that hint didn’t help me to identify the sequence of numbers. Do you know what the sequence is? **

A recent meeting of the advisory committee was held at a private school in NYC, and a number of parents were waiting in the hallway when our meeting ended. As I walked by, one of the parents stood up quickly, and I accidentally brushed against her. “Oh, I’m sorry, ma’am,” I said. She turned to look at me, and I looked back. First, I noticed how tall she was. Then I noticed something else. “Oh,” I said, “you’re Brooke Shields.” Turns out her kids go to this school. She smiled politely at my recognition.

She was dressed in casual clothes, and she was just there to pick up her kids. I didn’t want to be a nuisance, so I just said, “Have a great day.”

How cool is that? Go to a math meeting, meet a celebrity! And one I had a crush on when I was 13, no less!

** The sequence is the number of sides for *constructible polygons*, which are regular polygons that can be drawn with a straightedge and compass.

### Dissections are Fun, No Matter How You Slice It

Some folks have referred to me as “a real cut‑up.” Of course, those folks are generally septuagenarians who still use words coined in the mid‑1800’s. But I take it as a compliment, perhaps because I’m a fan of dissection problems.

Speaking of dissection, here’s one of my favorite quotes about jokes, which comes from E. B. White:

Humor can be dissected as a frog can — but the thing dies in the process, and the innards are discouraging to any but the pure scientific mind.

Like jokes and amphibians, geometric figures can be dissected, too. The following are three of my favorite dissection problems.

My all-time favorite is the Haberdasher Problem, originally proposed by H. E. Dudeney in 1902. It’s my favorite because of its simplicity — it can be described with just a few words, and it asks the would‑be solver to turn the simplest of all polygons, an equilateral triangle, into a square:

With three cuts, dissect an equilateral triangle so that the pieces can be rearranged to form a square.

The second isn’t really a problem to be solved. Instead, it’s a dissection that offers a surprising result. (At least, I was surprised the first time I saw it.)

Remove a tetrahedron of edge length 1 unit from each vertex of a larger tetrahedron with edge length 2 units. In total, remove four tetrahedra, one from each vertex. The image below shows one of the tetrahedra being removed. What is the shape of the solid that remains?

Finally, here’s an original dissection problem.

Form a pentagon by placing an isosceles triangle with height 1 unit on top of a square with side length 2 units. Then find a point

Palong the perimeter of the pentagon such that whenPis connected to two vertices, the three pieces into which the pentagon is divided can be rearranged to form a square. Find the area of the largest piece.

As usual, these problems are presented without answers or solutions. More fun can be had if you solve them yourself.

### Fair and Square Solution

A few days ago, I posted the following problem:

Three points are randomly chosen along the perimeter of a square. What is the probability that the center of the square will be contained within the triangle formed by these three points?

In that post, I mentioned that I had an elegant solution. If you want to think about the problem a little before seeing my solution, check out the Three Points page at the MJ4MF website. I created a really cool Excel file that allows you to explore 100 cases at a time. I think it’s worth a look.

But if you’re not interested in any of that, here’s my solution…

Without loss of generality, choose any two points along the perimeter. Call them A and B. Then, construct segments AD and BC that pass through point P, the center of the square.

This gives four cases:

- If A and B are the two chosen points, then the third vertex must be chosen along segment CD to form a triangle that contains point P.
- If B and D are the two chosen points, then the third vertex must be chosen along segment CA.
- If D and C are the two chosen points, then the third vertex must be chosen along segment AB.
- If C and A are the two chosen points, then the third vertex must be chosen along segment BD.

What we’ve now done is selected two sets of points in **four** different ways, and the portions along the segments where the third point could be chosen collectively comprise the **entire perimeter**. Consequently, the probability that a point along the perimeter is 1, but this is divided among 4 cases, so the probability that a randomly selected triangle will contain the center is **1/4**.

As a footnote, points A and B were chosen arbitrarily, so there is no loss of generality.