Our last Geometry Problem of the Week, Voulez Vous des Voussoirs?, deals with building arches using trapezoidal blocks. Maybe you’ve had a chance to build such an arch at a museum. I’ve done it most recently with my sister’s family at the Minnesota Children’s Museum. In our problem, if you’re told the angle measure of the obtuse angles of the “trapezoid”, how can you figure out how many identical blocks are needed to build a semi-circular arch?

What is most interesting to me about this problem is that different people seem to “see” it in different ways. That is, they get one particular model in their head of the situation. When I first solved it, I “saw” it like this:

A few students saw it that way, though it wasn’t the most common method we saw. Here’s an excerpt from one solution:

Renuka D

I knew if the obtuse angles were 96 degrees, the supplement would be 84. Because of symmetry, the two sides of the voussoir must meet at the center of the semi-circle. In this way a triangle is formed, so therefore the angle at the center of the circle must be 180 – (84+84).

Here’s the second idea I “saw” when I solved this:

This was the basis of the most common method that students used this week, and is explained in excerpts from two solutions:

Gavin T, Highlands Elementary School

I knew that if all the angles were 90 degree angles, the arch wouldn’t be an arch, it would go straight up. 96 degrees meant that each angle angled the arch 6 more degrees.

Jed M, Waterford Elementary School

If there is a 6° increase on each angle. (I know that because there 96° angles, and if it was a 90° angle it would go straight up.) And there’s two angles on each voussoir. So theres a total of a 12° increase. I’ll use math sentences to get a total of 180 and what ever number times twelve equals 180 is my answer.

Then there’s the image conjured up by a group of students from Conners Emerson School in Maine. They made it a problem about angles of regular polygons. That had never occurred to me, so it was exciting to get their submission. Here’s my version of what they “saw”:

Here are some excerpts from their solution, including a hint and their picture:

Xingyao C, Tarzan M, and Tom G, Conners Emerson School

We drew a diagram of two of the voussoirs adjacent to each other. If the voussoirs make it all the way around, they would form a regular polygon.

The two adjacent acute angles of the trapezoids make an interior angle of the regular poylgon.

Hint: Try not to look at the voussoirs as solitary pieces, but as part of a convex polygon. You are not trying to find the information of each individual voussoir, but the information of the arch made by many.

(I can’t help but point out that they named their picture “French Words in Math”, which I think is awesome!

I wonder how you think your students “saw” this situation. Was one solution method more common, or was there a variety of models used in your classroom? How did you see the situation? Please let us know!

Some Voulez Vous de Voussoirs? links in case you are interested:

• The problem [requires a Math Forum PoW Membership].
• Information about accessing Voulez Vous de Voussoirs? (and a selection of all our PoWs) for 21 days with a free Math Forum trial account.
• Information about becoming a Math Forum Problems of the Week Member. Consider starting with a $25 membership, which gives you access to all of this year’s Current PoWs — and now you can create 36 student logins as well!

The last Math Fundamentals Problem of the Week was a puzzle of sorts. You’re told there are three chips with a number on each side. The six numbers are consecutive one digit positive numbers. One way to throw the chips gives you 6, 7, and 8, for a sum of 21. You’re also given four other sums between 16 and 23 that are possible. Aside from adding three numbers together, there isn’t any other math concept that’s central to this problem. It isn’t about “fractions” or “proportional reasoning” or “measurement”. It’s about understanding what’s going on, and coming up with a method to figure out what numbers are on the other sides of the chips.

Since solving a problem like this often lends itself to running around in circles for a bit before you hit upon the answer, the resulting submissions are usually of two different types. The first is the solution that states the answer, and then confirms that the answer is correct. Here’s an example:

On the back of the 6 there is a 5, on the back of the 8 there is a 4, and on the back of the 7 there is a 9

for 16 i added: 7 plus 5 plus 4.

for 17 i added: 6 plus 7 plus 4.

for 20 i added: 6 plus 5 plus 9

for 23 i added: 9 plus 6 plus 8

That example doesn’t give me any idea how the student approached the problem, what sorts of things they tried or did, whether they made any mistakes and found them, or anything that seems systematic and efficient. We could do some work, using the sums that they provided, to figure out the correct combinations. But the student doesn’t provide any insight into the process they used to find the answer.

It’s not uncommon to see a lot of submissions that confirm results. Sometimes it’s hard to keep track of the different paths you traversed on your way to the answer. Good recordkeeping is a must. I have to say, however, that reading the solutions for this problem, I was excited to see so many students telling the “story” of how they solved the problem. Most used some form of the Guess and Check strategy. Some used Logical Reasoning. Here are a couple of examples:

Told 6+7+8=21. Number are consecutive, and to add to 23, one number has to be 9, for 6+9+8. 6 and 8 stay- 7 and 9 are on the same chip. 4 and 5 are the other numbers. To add to 16 you have 4+5+7. 17 needs 4+6+7. 5 and 6 have to be on the same chip. 8 and 4 are the remaining numbers and are on the same chip.

First I thought, since the toss of 23 is larger than the 6, 7, 8 toss, it must contain a number greater than 8. Therefore 9 is needed. The toss must be 9, 8, 6 because no other combination gives 23. So the 7 and 9 are on the same chip. This also means the other two numbers must be 4 and 5. Next I thought the toss of 16 must be made of 4, 5, 7. Then, the toss of 17 couldn’t be 4, 5, 8 because it is not possible since the 7-9 chip is missing.
So 4 must be on the 8 chip to have all 3 chips in the toss. This gives the 3 chips: 5/6, 4/8, 7/9.
Then I concluded that the 20 roll is possible with 5, 7, 8 so it all checked out.

However they did it, we really love reading stories about students’ problem solving adventures. How do you support your students in telling this story? Is it something that gets reported orally in the classroom a lot (which is great practice for eventually writing it down)? Do they read aloud to others? Do you just keep prodding them to say more about their process? We’d love to hear from you, since all the kids submitting to this problem didn’t do a good job by luck or accident!

Some Let the Chips Fall… links in case you are interested:

• The problem [requires a Math Forum PoW Membership].
• Information about accessing Let the Chips Fall… (and a selection of all our PoWs) for 21 days with a free Math Forum trial account.
• Information about becoming a Math Forum Problems of the Week Member. Consider starting with a $25 membership, which gives you access to all of this year’s Current PoWs — and now you can create 36 student logins as well!

Join the John Ehret Patriots in New Orleans, Lousiana, USA, in Noticing and Wondering about this Problem of the Week!

The Hopi Indians invented Totolospi, a game of chance. The game is played with three cane dice, a counting board, and a counter for each player. Each cane die can land round side up (r) or flat side up (f).

The moves of the game are determined by tossing the three cane dice with these rules:

Toss	Move
three round sides up (rrr)	player advances 2 lines
three flat sides up (fff)	player advances 1 line
any other toss of the three cane dice	player doesn't advance

This week’s Pre-Algebra Problem of the Week, “Happy New Year Wish,” was so much fun! It’s based on the true wondering that our colleague Suzanne had when she wished her son, Specialist Lee Alejandre, “Happy New Year!” while he was stationed in Seoul, South Korea. She wondered why the time was 14 hours earlier in Seoul, and what it had to do with the longitudes of Philadelphia and Seoul.

I had so much fun reading students’ work as they connected their understanding of math to their thoughts about time zones, globes, and longitudes. There were many different solution methods, and students talked a lot about how they got started on the problem… whether it was talking to friends, getting out a globe, drawing a picture, Googling, or thinking of a simpler related problem. Here are some of my favorite quotes! I wonder if this problem was extra “juicy” because it was about a real puzzle…

Student eleven from Caughlin Ranch ES brought up an important point: what happens if we thought about going east from Seoul to Philadelphia, or west from Philadelphia to Seoul? Why can’t we base our calculations on the 157º difference. Here’s what Student eleven had to say, “i also remembered a time when a book was answering a question involving the international dateline. this told me that the same thing would work going west but you’d go +1day. so goin east would be less complicated (i assumed).”

Julia R. from Birch Wathen Lenox School helped me think about how to decide which way to round in this problem when she said, “Although the numbers werent exact, 210 degrees meant that the time was passing through the 14th hour of longitude, and aproaching the 15th. It was closer to the 14th hour though.” Did you think about why, if 203º of difference is 13.53333… hours apart, we round to 14 hours instead of 13? Did looking at a globe help you decide? Read More→

I was going to write this blog post about moving from guess and check to an algebraic solution method, and why you might want to use algebra.

Then I was reading through some of the Guess and Check submissions we got, and thinking, “wow, that looked pretty effortless.” One student, Jack M. from Rosemont School of the Holy Child, pointed out that the solution has to be a multiple of 3 between 20 and 40. If you start your guess and check with some of that reasoning, you know you won’t have to do very many guesses. And if you’re good at adjusting a guess up or down based on your results, you can get the answer pretty quickly. Jack only needed two guesses: he started with the first multiple of 3 above 20, which is 21, and got to 24 on his second guess!

Rashmi R. from West Woods Upper Elementary School got the answer in 5 systematic guesses (he didn’t focus on the multiple of 3 idea, but was still very efficient). He guessed 10 CDs were sold, then 20, then 30, then 25, then 24. Each time he used the data from the previous guess to think about if he needed to increase his guess or decrease to something between the previous two guesses.

Adam S. from Highlands Elementary School may have had a lucky first guess or he may have thought hard about a reasonable starting number… he doesn’t say. Either way, he started with 25 as first guess, realized it was too high, adjusted to 23 for his second guess, realized it was too low, and got to 24 on his third guess. Take a look at his work, below:

Read More→