If you asked a mathematician how to make an origami coin purse for Chinese New Year, you might receive the following instructions:
- Rotate a square piece of paper 45°.
- Reflect an isosceles right triangle about a diagonal of the square so that congruent isosceles right triangles coincide.
- Reflect the lower left vertex of the isosceles right triangle so it coincides with the opposite leg, such that the base of the isosceles right triangle and one side of the newly formed isosceles acute triangle are parallel.
- Reflect the other vertex of the isosceles right triangle to coincide with a vertex of the isosceles triangle formed in the previous step.
- Fold down a smaller isosceles right triangle from the top.
- Fill the coin purse with various denominations of currency, such as a $2 bill, a $1 gold coin, a quarter, or a penny.
Don’t like those directions? Perhaps these pictures will help.
The Common Core State Standards for Mathematics (CCSSM) have added rigor to geometry by requiring students to prove no fewer than a dozen different theorems. Further, the progressions document formulated by Dr. Wu suggests that a high school geometry course should use a transformational approach, basing a majority of the foundational axioms on rotations, translations, and reflections.
What struck me about the origami coin purse was the myriad geometry concepts that could be investigated during this simple paper-folding activity. Certainly, the idea to use origami in geometry class in nothing new, but this one seems to have a particularly high return on investment. Creating the coin purse requires only four folds, and though perhaps I lied by claiming forty theorems in the title, the number of axioms, definitions, and theorems that can be explored is fairly high.
For instance, the fold in Step 3 requires parallel lines, and the resulting figures can be used to demonstrate and prove that parallel lines cut by a transversal result in alternate interior angles that are congruent.
No fewer than four similar isosceles right triangles are formed, and many congruent isosceles right triangles result.
The folded triangle in the image above is an isosceles acute triangle; prove that it’s isosceles. And if you unfold it from its current location, the triangle’s previous location and the unfolded triangle together form a parallelogram; prove it.
The final configuration contains a pair of congruent right triangles; you can use side-angle-side to prove that they’re congruent.
But maybe you’re not into proving things. That’s all right. The following are all of the geometric terms that can be used when describing some facet of this magnificent creation.
- Right Angle
And the following theorems and axioms can be proven or exemplified with this activity.
- Sum of angles of a triangle is 180°.
- Alternate interior angles formed by a transversal are congruent.
- Any quadrilateral with four congruent sides is a parallelogram.
- Congruent angles are formed by an angle bisector.
- Triangles with two sides and the included angle congruent are congruent.
There is certainly more to be found. What do you see?
I don’t know where you live, or how much snow you’ve gotten, or whether your kids have been out of school for multiple days. But here in Falls Church, VA, it’s Thursday, January 28 — five full days after snow stopped falling from Winter Storm Jonas — and our schools are still closed.
My sons lounge around in their pajamas all day, only getting off the couch to interrupt my work-from-home day and ask for macaroni and cheese. It’s starting to feel like we’ve had two eight-year-old brothers-in-law take up residence.
That’s why I’ve used data from Snowzilla to create a series of math activities. Today’s assignment is for them to complete the following and not bother me till they’re done. Feel free to use any of these with the youngsters in your life, whether they’re your biological offspring from whom you need a break at home or your charges in a classroom who might enjoy the challenge.
What do geometry teachers do in a blizzard?
Make snow angles.
Schools have had a record number of snow days. At this rate, the only math kids are doing is how many glasses of wine their mom drinks before 2 p.m. – Jimmy Fallon
1. Chris Christopher, a macroeconomist at IHS Global Insight, estimated that Jonas’s economic impact would be somewhere between $500 million and $1 billion.
a. Write both of those numbers in the form 2m × 5n.
b. If all values within the range are equally likely, what is the probability that the impact will be greater than $800 million?
c. If all values within the range are normally distributed, what is the probability that the impact will be greater than $800 million?
Bonus. Can you think of another occupation where it’s appropriate to state a prediction in which the upper end of the range is double the lower end?
2. In the Washington, DC, area, the average weekly sales in a typical supermarket is about $10 per square foot. In the two days leading up to Jonas, traffic to brick-and-mortar stores was 7.5 percent higher than usual. The graph below, based on national averages, shows the percent of weekly shoppers at grocery stores each day of the week.
Putting all this information together, as well any other data that you can find online, draw a graph that approximates sales at a typical grocery store in Washington, DC, for the month of January.
3. Virginia Governor Terry McAuliffe estimated that snow removal costs the commonwealth $2 million to $3 million per hour. Estimate the total cost for Virginia to clean up Jonas’s mess.
4. According to City Comptroller Scott Stringer, the cost of snow removal in New York City is approximately $1.8 million per inch. Estimate the total cost for New York City to remove the snow from Jonas.
5. The estimate above is an average for 2003 to 2014. The two graphs below show the snowfall totals and snow removal costs for those 12 years. Which years had the highest and lowest cost-per-inch? (Click each image to enlarge.)
Yes, I know that I just posted some Math Problems for 2016 on December 19.
But I’ve decided to post some more for a variety of reasons:
- 2016 is cool.
- It’s a triangular number.
- It has lots of factors. (I’d tell you exactly how many, except that’s one of the problems below.)
- After writing the problems for that previous post, I just couldn’t control myself.
- It’s my blog, and I can do what I want.
People who write math problems for competitions (like me) love to be cheeky and include the year number in a problem, especially when any sufficiently large number will do. When the year number is critical to the success of a problem, well, that’s just a bonus. With that in mind, there are 16 problems below, each of which includes the number 2016.
A fully formatted version of these problems, complete with answer key, extensions, and solutions, is available for purchase through the link below:
Enjoy, and happy new year!
- What is the sum of 2 + 4 + 6 + 8 + ··· + 2016?
- Using only common mathematical symbols and the digits 2, 0, 1, and 6, make an expression that is exactly equal to 100.
- Find a fraction with the following decimal equivalent.
- How many positive integer factors does 2016 have?
- What is the value of n if 1 + 2 + 3 + ··· + n = 2016?
- Find 16 consecutive odd numbers that add up to 2016.
- Create a 4 × 4 magic square in which the sum of each row, column, and diagonal is 2016.
- Find a string of two or more consecutive integers for which the sum is 2016. How many such strings exist?
- What is the value of the following series?
- What is the units digit of 22016?
- Some people attend a party, and everyone shakes everyone else’s hand. A total of 2016 handshakes occurred. How many people were at the party?
- What is the value of the following expression, if x + 1/x = 2?
- A number of distinct points were placed along the circumference of a circle. Each point was then connected to every other point, and a total of 2016 segments were formed. How many points were placed on the circle?
- Let A = 1, B = 2, C = 3, …, Z = 26. Find a word for which the product of the letters is 2016. (This one may look familiar.)
- Each dimension of a rectangular box is an integer number of inches. The volume of the box is 2016 in3. What is the least possible surface area of the box?
- What is the maximum possible product for a set of positive integers that have a sum of 2016?
UPDATE: Bonus Material!
Special thanks to my friend Harold Reiter, who created the following 2016 problems for use as MathCounts practice:
- What is the smallest number N such that the product of the digits of N is 2016?
- What is the sum of the divisors of 2016?
- What is the product of the divisors of 2016? Express your answer as a product of prime numbers.
- Solve the following equation:
- What is the binary representation of 2016?
- What is the base-4 representation of 2016?
- What is the base-8 representation of 2016?
With one week left in the NFL season, Dan Graziano had this to say about the Indianpolis Colts’ chances of making the playoffs:
Indianapolis can still win a third straight AFC South title. Really, it can. All it needs is to win and then have the Texans, Bengals, Chargers, Jets, Saints, Chiefs, Patriots and Browns all lose. The league will throw in the partridge in a pear tree.
If the Colts win and the Houston Texans lose, both would be 8-8, and the first four tiebreakers for deciding which team makes the playoffs – record against one another, record against divisional opponents, record against common opponents, and record within the conference – would not be enough to decide who makes the cut.
It then comes down to strength of victory and strength of schedule. And for things to play out in the Colts’ favor, a lot of things have to go their way.
Fox Sports referred to this as long shot, comparing it to a recent win by a horse who was 200-to-1:
Sure, going 9 for 9 here looks dim, but long shots come in every once in a while.
Are the Colts’ odds as good as that horse’s? Seems not.
Using a simplistic model, assume that each of the nine necessary outcomes are equally likely. That alone would put the Colts’ odds at 511-to-1. (Since 29 = 512.)
But it’s not that simple. The following chart from 538.com gives the probability of each team winning their game this weekend:
The good news is that the Colts have an 81% chance of winning their game against Tennessee. The bad news is that it seems unlikely that any of the Texans, Bengals, Chiefs, or Patriots will lose, let alone all four of them. So putting all those numbers together, the Colts’ chances of making the playoffs are:
0.81 × 0.20 × 0.22 × 0.84 × 0.54 × 0.69 × 0.15 × 0.18 × 0.77 = 0.0002 = 0.02%
or, more precisely, about 4,311-to-1. That’s more than a long shot; that’s an extended-to-an-unfathomable-distance shot.
The Colts are in the unenviable position of Lloyd Christmas in Dumb and Dumber, “So, you’re telling me there’s a chance…”
“What homework do you have to do tonight?”
I ask my sons this question daily, when I’m trying to determine if they’ll need to spend the evening doing word study or completing a math worksheet, or if we’ll instead be able to waste our time watching The Muppets or, perhaps, pulling up the animated version of Bob and Doug Mackenzie’s 12 Days of Christmas on Dailymotion.
When I asked this question last night, though, the answer was surprising:
We have to do our reading, but we already completed your math problem.
My problem? I had no idea what this meant. So they explained:
It’s not a problem you gave us. It’s one we got from [our teacher], and it says, “This problem was written by Patrick Vennebush.”
I was puzzled, but then it dawned on me. I asked, “Does it have a monkey at the top with the word BrainTEASERS?”
“It’s about the word CAT.”
I knew the problem immediately. It’s the Product Value 60 brainteaser from Illuminations:
Assign each letter a value equal to its position in the alphabet (A = 1, B = 2, C = 3, …). Then find the product value of a word by multiplying the values together. For example, CAT has a product value of 60, because C = 3, A = 1, T = 20, and 3 × 1 × 20 = 60.
How many other words can you find with a product value of 60?
As it turns out, there are 14 other words with a product value of 60. Don’t feel bad if you can’t find them all; while they’re all allowed in Scrabble™, the average person won’t recognize half of them.
You can see the full list and some definitions in this problem and solution PDF.
This problem resurfaced at the perfect time.
With 2016 just around the corner, no doubt many math teachers will present the following problem to students after winter break:
Find a mathematical expression for every whole number from 0 to 100, using only common mathematical symbols and the digits 2, 0, 1, and 6. (No other digits are allowed.)
And that’s not a bad problem. It gets even better if you require the digits to be used in order. For instance, you could make:
- 2 = 20 + 16
- 9 = 2 + 0 + 1 + 6
- 36 = (2 + 0 + 1)! × 6
But that problem is a bit played out. I’ve seen it used in classrooms every year since… well, since I used it in my classroom in 1995.
So here are two versions of a problem — the first one being for younger folks — using the year and based on the Product Value 60 problem above:
How many words can you find with a product value of 16?
How many words can you find with a product value of 2016?
There are 5 words that have a product value of 16 and 12 words that have a product value of 2016 (spoiler: those links will take you to images of the answers). As above, you may not recognize all of the words on those lists, but some will definitely be familiar.
I recently received an email from Akram in Lebanon who saw the Rebus Quiz on the MJ4MF website and suggested the following rebus:
The answer is below, but I’ll give you two hints.
The first is that you might say the same thing to a calculus student who spent 4 hours completing an integration problem only to forget the constant.
The second hint is that Akram’s suggested rebus reminds me of this one, which you may have seen before and is very appropriate for the holidays:
The answer to this one is below, also.
But Akram’s email convinced me that it was time to create a few more, so here are five all-new rebuses (or is that rebi?) for you to enjoy.
To put a little space between the rebuses and the answers below, here’s one more rebus from Sandy Wood:
SPOILER ALERT… Answers Below
ENTURY: Long time, no C (see)
ABCDEFGHIJKMNOPQRSTUVWXYZ: No L (noel)
- A + 6 backwards = AXIS
- SINE + double U = SINEW
- ONION + PI in the middle = OPINION
- Dr. OZ + ONE = OZONE
- STUN + TWO + MAN = STUNTWOMAN
During yesterday’s NPR Sunday Puzzle, puzzlemaster Will Shortz presented the following challenge:
I’m going to give you some five-letter words. For each one, change the middle letter to two new letters to get a familiar six-letter word. For example, if I said FROND, F-R-O-N-D, you’d say FRIEND, because you’d change the O in the middle to I-E.
He then presented these nine words:
For those of you who don’t know who Will Shortz is, you have something in common with detective Jake Peralta from Brooklyn Nine-Nine:
The puzzle was fun. But what was more fun was the conversation that our family had about it. After the third word, Alex announced, “This shouldn’t be that hard. There are only 676 possible combinations.”
What he meant is that there are 26 × 26 = 676 possible two-letter combinations, which is true.
He continued, “But you can probably stop at 675, because Z-Z is pretty unlikely.”
I smiled. He had chosen to exclude Z-Z but not Q-K or J-X or V-P.
Yet his statement struck me as a challenge. Is there a five-letter word where the middle letter could be replaced by Z-Z to make a six-letter word? Indeed, there are several:
- BUSED or BUSES
- CONED or CONES
- FUMED or FUMES
- RAVED or RAVES
- ROVED or ROVER or ROVES
- WINED or WINES
None of them are perfect, though, because Z-Z is not a unique answer. For instance, ROVER could become ROBBER, ROCKER, ROMPER, ROSTER, or ROUTER, and most puzzle solvers would surely think of one of those words before arriving at ROZZER (British slang for a police officer).
From the list above, the best option is probably GUILE, for two reasons. First, stumbling upon GUZZLE as the answer seems at least as likely as the alternatives GUGGLE, GURGLE, and GUTTLE. Second, the five-letter hint has only one syllable, but the answer has two, and such a shift makes the puzzle just a little more difficult.
But while Alex had reduced the field of possibilities to 675, the truth is that the number was even lower. The puzzle states that one letter should be “changed to two new letters,” which implies that there are only 25 × 25 = 625 possibilities. Although that cuts the number by 7.5%, it doesn’t help much… no one wants to check all of them one-by-one to find the answer.
When Will Shortz presented DEITY, the on-air contestant was stumped. So Will provided some help:
I’ll give you a tiny, tiny hint. The two letters are consonant, vowel.
Alex scrunched up his brow. “That’s not much of a hint,” he declared.
Ah, but it is — if you’re using brute force. To check every possibility, this reduces the number from 625 to just 21 × 5 = 105, which is an 80% reduction.
Still, Alex is correct. The heuristic for solving this type of puzzle is not to check every possibility. Rather, it’s to think of the word as DE _ _ TY, and then check your mental dictionary for words that fit the pattern. It may help to know that the answer isn’t two consonants, but most puzzle solvers would have suspected as much from the outset. In the English language, only SOVEREIGNTY, THIRSTY, and BLOODTHIRSTY end with two consonants followed by TY.
Below are five-letter math words for which the middle letter can be changed to two new letters to form a six-letter word. (Note that the answers aren’t necessarily mathy.)
DIGIT :: DI _ _ IT (unique)
POINT :: PO _ _ NT
FOCUS :: FO _ _ US
MODEL :: MO _ _ EL (unique)
POWER :: PO _ _ ER
RANGE :: RA _ _ GE (unique)
SOLID :: SO _ _ ID (unique)
SPEED :: SP _ _ ED
And below, your challenge is reversed: Find the five-letter word that was changed to form a six-letter math word.
CO _ EX :: CONVEX (unique)
LI _ AR :: LINEAR (unique)
OR _ IN :: ORIGIN
RA _ AN :: RADIAN (unique)
SE _ ES :: SERIES
SP _ RE :: SPHERE