During the February 26th MoMath Masters Tournament, @MoMath1 tweeted, “No googling – how many sides on an enneagon?” We thought, “Hey! We know enneagons!” If you don’t, maybe this scenario from a problem we first used in 1998 will give you some hints (as well as some ideas for something you could do with one).

Extend the sides AB and ED of the regular enneagon ABCDEFGHI until they intersect.

## Baseball Cards

Third grade students at Hanover Street School made this awesome video as a Free Scenario based on the Math Forum problem called “Baseball Cards.” We are so excited to share their video!

The students’ video is based on the Math Forum Baseball Cards Scenario [PDF]

I’ve been reading a lot lately about the idea of a “modeling curriculum.” Not as in *America’s Next Top Model *and also not as in the teacher models the thinking and the student learns from watching and trying it themselves. A modeling approach to teaching science and math means that the students work together to develop better and better conceptual models to explain situations. So in physics, you might roll two objects down a ramp and try to make a mathematical model to describe what was going on. At first you might include the weight of the balls in your model, but then you might observe that two objects with different weights behave the same, and so your model would change based on new data and new understanding.

Some of the studies of this kind of teaching show us that students come into situations with models already in their heads — they already have ideas about how balls fall, for example. Their models might not be the most accurate or easiest to use, and so as they encounter new situations and new demands, they change their models. While that’s happening, students might use lots of different competing ideas at once. One minute the same kid will go from making really accurate predictions about two balls of different weights rolling down an incline, but then say that gravity will make a bowling ball fall faster than a beach ball.

This week’s AlgPoW, Filling Glasses, asked students to match graphs of water level vs. time of glasses being filled at a steady rate, to pictures of the glasses. Students used many different models for thinking about the problem:

- Try to match the shape of the graph to the shape of the glass (e.g. count the wavy parts, look for straight graphs for straight glasses).
- Relate the skinniness of the glass to speed.
- Relate the skinniness of the glass to the steepness of the graph.
- Relate the height of the glass as a whole to the maximum height reached in the graph.
- Relate the skinniness of the graph to speed at which the glass fills and the speed at which the glass fills to the steepness of the graph.

What was most interesting, though, was the students who used different strategies at different moments. Students who are in the middle of learning often switch models based on small details or when a problem seems easier or harder for some reason.

Like this:

For this problem, you have to really visualize the glasses and their shape.

First, I looked at glass A. It starts out skinny for a tiny bit, then there is a huge bulge before it is a little skinnier. So the height would rise quickly for the shortest amount of time, then go slower, then finally go a little faster. I visualized the graph to be a slightly zigzaggy line that was not too tall. Graph 4 did not have any zigzags, and graphs 2 and 3 went too high. So, graph 1 matched with glass A.Glass B is like a funnel, starting skinny and getting wider and wider as the top draws nearer. So the height would rise quickly at first and get slower and slower. Since there are no bulges in glass 2, the graph it matched up to would have to be zigzag-free. And the only graph without zigzags is graph 4.

Finally, glass C starts skinny, gets wider, gets skinnier, and then gets wider. The water will go fast at first, then slower, then faster, then slower. Graphs 2 and 3 are very similar, but only graph 3 starts out fast.

The student sometimes is looking for zig-zags, basically matching the shape of the glass to the shape of the graph. But in the case where there ar

e two possible zig-zag graphs that could match one glass, the student switches to a (more robust?) model of thinking about the width -> speed relationship (and maybe implying a speed -> steepness relationship?).

Or this:

Glass A= First of all glass A is the shortest so the line on the graph would be less steep. Also, since the glass is kind of round, at first the water would pour fast then gradually pour slower then after you get to the middle the water would gradually pour faster.

The thinking about how steepness relates to the shortness of the glass seems like a very different way of thinking about steepness than the speed idea that she uses after “Also,…”

Or finally:

i figured this out becauause if you look at the glasses and the graphs. the arches in the graphs are like the glasses when get bigger because you need to have more water and then it would fill it up.

There’s the shape kind of thinking there: “the arches in the graphs are like the glasses” but also some idea of the change in width of the glass affecting how it fills up.

**Some “Filling Glasses” links** in case you are interested:

- The problem [requires a Math Forum PoW Membership].
- Information about accessing “Filling Glasses″ (and all our current PoWs) for two weeks with a free Math Forum trial account.
- Information about becoming a Math Forum Problems of the Week Member. Compare prices – consider starting with a $25 membership giving you access to all of this year’s Current PoWs — and now you can create 36 student logins as well!

Last year brought us the unveiling of the Primary Problem of the Week, a series of PoWs geared towards the youngest problem solvers. This year, we’re focusing on older students, students who have studied math beyond Algebra I and Geometry. We will be publishing 20 problems from our Trig & Calc library to make them available to all teachers with a Current PoW membership. Each problem will have links to enhanced teacher materials (strategy alignments, Online Resource Pages, Scenario-Only versions of the problems, and Teacher Packets including Common Core alignments).

We’re excited to be able to offer these problems to our Current PoW members so that teachers of higher-level math can be part of the Current PoW community. As the name suggests, most of the problems can be solved using techniques from Trigonometry or Calculus. However, many of the problems can be solved in multiple ways: with right-triangle trig that students may have learned in early grades; with algebraic techniques and software; or they draw on content areas like Discrete Math or Probability. So these problems can also be extra challenges for students who aren’t yet in the Trigonometry or Calculus class.

This year, we don’t plan to feature highlighted solutions for the Trig & Calculus PoW on the PoW site (teachers can always see sample successful solutions with different strategies in the Teacher Packet). If, however, we do get interesting submissions, we’ll certainly be blogging about them here! And if we get to a consistent level of submissions, we’ll be excited to have highlighted solutions from the Trig & Calculus PoW next year!

So we’d love to have you check out the current Trig & Calculus PoW, “Building Boxes,” a PoW that can be approached by older students using derivatives or by middle-school students through careful tables and virtual manipulatives. View the Online Resources Page in the “blue box” on the PoW for a link to an applet that will help students from middle school through calculus make sense of this problem.

by Max

March
16th,
2012

This week, we asked Pre-Algebra students to solve a problem that turned out to be really tough. Students were asked to think about scoring in a game used to practice place value skills in the context of scientific notation. We used some examples to illustrate how the scoring worked, and many students used the examples to start teaching themselves place value in scientific notation.

It was pretty awesome to have a window into how students interpreted the examples we gave. Some students were able to make sense of what was happening. Other students got correct answers, but some of their words made me wonder if they were getting the right answer without really understanding what was going on. And some students saw things in the examples that I never would have thought of… that might lead to problems down the road! The whole experience could be a microcosm of what students see when teachers work out examples on the board.

Here are the examples we gave:

If a card contains the number 3.38 x 10^14 and Mrs. Nasamy picks an 8, the card is worth 8 x 10^12 points. If she picks a 3, it’s worth 3.3 x 10^14 points since 3 appears in two places, you must add both values:(3 x 10^14) + (3 x 10^13) = (3.3 x 10^14) points

We include this example so you can think about how the game is played when the digit pulled appears in more than one place; note that the process would be the same if the digit appeared on two different cards.

And here are some ways students interpreted those examples:

Student 1 thinks about how the exponent and place value of the target digit are manipulated together:

When the 3.38×10^14 card was there the next time they showed it had gone from 3.38×10^14 to 8×10^12. It was like this because you cross out everything but the number pulled. If the number pulled is the first in back of the decimal the exponent needs to take 1 away. If it is the second back you do the same thing but take two away from the exponent.

Student 2 has an AHA! moment relating to the distributive property:

first off, i had to do a lot of time figuring out what the question is. it took me a while, but an AHA! was when i realized that these cards were like three #’s in sci. notation mashed together for example, (1.92×10^9) is (1×10^9)+(9×10^8)+(2×10^7). and if a 2 is pulled, your points are (2×10^7).

Student 3 provides a third example to check his/her understanding:

my interpretation of the problem is this. The teacher pick a card between 1 and 9. Whatever number she draws, you find the number in your scientific notation. for example, if she drew a 7, and one of your notations was 7.37 X 10^

^{3}, it would turn into 7.07 X 10^^{3}. then you would solve the exponent, which is 10^^{3}, and that is 1,000. Now your problem is simply 7.07 X 1,000. When you solve that simple multiplication problem, you get 7,070. That is your answer for that card.

Student 4 applies the examples to the main problem, finding the point score if Ms. Nasamy pulls an 9 and you have cards with 1.92 x 10^9 and 6.59 x 10^8. I am not sure if Student 4 has a misconception or not, because they do eventually come to the correct answer, however, I suspect something’s wrong after very careful reading:

I…started with 1.92*10^9. I then need to cross out the place values that aren’t 9, like the example showed us. Now the problem is 0.9*10^10. The exponent changed because I took the 0.02 place value away. Now, since there are 10 zeros instead of 9, the exponent changes to 10. To solve the problem, I first calculated 10^10, which is 10,000,000,000. Then I did 10,000,000,000*9, which is 90,000,000,000. Then I did the last one which is 6.59*10^8. Next I used the same process for the last card and changed the problem to 0.09*10^8. The exponent didn’t change because the 0.09 is the last place value there. After that, I did the same process as the first card and and solved the problem. I ended up with 900,000,000. Then I added the points together and got 90,900,000,000. I then converted that number into a decimal. I did that by counting the number of place value zeros and the turning that number into the exponent next to ten. I got 9.09*10^8.

A middle-school math teacher, Ms. Alcala, shared the activity, “My Favorite No” on the Teacher Channel. I highly recommend watching the video. The idea is to collect work from students and then share, anonymously, one incorrect solution that you really like. You might like it because it’s a different approach or because of how much was correct about it or because it’s a common mistake. The idea is to share some work that’s not perfect in a very positive, non-judgmental way, and engage students in thoughtfully critiquing the work. Students get to work on the mathematical practice, “Construct Viable Arguments and Critique the Reasoning of Others” while also sharing and strengthening their content knowledge.

This week, reading student submissions to “Teeter Trio” I was struck by this awesome, but ultimately incorrect, submission:

Seesaw Balance POTW

A seesaw can balance with more than two people on it. The product of each person’sweightanddistance from the fulcrumcontributes to the balancing. If the sum of those products on one side equals the sum of the products on the other side, balance is achieved.

-As far as I know balance can be achieved when both sides are equal when the products are added in other words, wd=wd, when w is weight and d is distance from the fulcrum.

Shareef and his two little sisters, Marshay and Janeka, are playing on a seesaw. Shareef weighs 30 pounds more than Marshay and 35 pounds more than Janeka, so Shareef sits on one side to balance the two girls on the other.

-If Shareef is trying to balance the other two girls on the other side than we need to know everyone’s weights to fill in the equation of wd=wd.

Shareef=s

Marshay=m

Janeka=j

-Since Shareef is 30 pounds more than Marshay and 35 pounds more than Janeka than we have two different equations for her.

s=30+m

s=35+j

-Since we know this information we can find Marshay’s weight by putting these two equations together like so.

35+j=30+m Subtract 30 on both sides

5+j=m

-For Marshay’s weight we get m=5+j and Janeka’s weight can’t be found so it will just be represented by the letter j.

Shareef is sitting 6 feet from the fulcrum and Janeka is sitting 4 feet from the fulcrum. If the seesaw is balanced, find a function that expresses Marshay’s distance from the fulcrum in terms of her weight.

-Now that we know Janeka’s distance and Shareef’s distance from the fulcrum but we need to know Marshay’s distance from the fulcrum. Her distance will be represented by the letter d. now that we have all the pieces of information than we can fill in the equation of wd=wd. (The right side will be Shareef and the left will be Marshay and Janeka)

wd=wd

6(35+j)=d(5+j)+4j subtract 4j on both sides

6(35+j)-4j=d(5+j) divide (5+j) on both sides

6(35+j)-4j/5+j=d distribute the 6 and combine like terms

210+2j/5+j=d Simplify

2(105+j)/j+5=d

Marshay’s distance from the fulcrum is 2(105+j)/j+5=d.

To help you think about what might be incorrect, it’s important to know that in the follow-up you will be given *Marshay’s* weight and asked to help calculate how far from the fulcrum she should sit.

My questions to you are:

1) What do you find awesome about this submission?

2) How would you fix it *without redoing the problem*? Is there a way to tweak this answer to make it better match what was asked?

Some ** Teeter Trio** links in case you are interested:

- The problem [requires a Math Forum PoW Membership].
- Information about accessing
*Teeter Trio*(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!

In the Math Fundamentals problem *Frog Farming*, the goal is to make at least four different rectangular pens, each of which uses 36 meters of fence. Many students thought of this the same way I did, which was to consider half the necessary perimeter as the sum of two adjacent sides.

Rachel B, Seven Bridges Middle SchoolI know that the problem was finding perimeter. I know first you divide 36 by 2 and get 18. Then you find addends of 18 and they are the length and width. I added 12 plus 6 which equals 18 and 12 times 2 plus 6 times 2 equals 36.

Sarah G, Laurel SchoolFirst I decided come up with a length and width for a rectangle that would equal 18 because 18 is half of 36 and you have to multiply that number by two to get the perimeter. I decided on 2 and 16. I checked it by doing 16+16+2+2= 36. One could by length=16, width=2.

Rachel and Sarah and I were thinking about perimeter, in the context of this problem, like this:

Another way I thought of this was as 2(L + W). Hmm…..

Then I was mentoring a few students in this problem and noticed that they were thinking about the problem differently.

Ethan Z, Lorne Park Public SchoolI thought of a rectange which has 4 sides and 2 sides are equal and the other 2 sides are equal because Farmer Mead wants a rectangular pen that uses 36m of fencing. First I got the answer by thinking of 2 equal numbers that add up to less than 36. Then, the last 2 equal numbers are the difference of 36 to the first 2 equal numbers. That’s how I got all the numbers of the first question.

Emily G, Laurel SchoolI used 2 numbers, and doubled 1 number by two (ex. 6×2=12). 36-12 is 24. 24 is an even number that can be split into 2, which is 12 (ex. 24÷2=12.) 24+12 is 36!

Maybe because I had “seen” the problem differently, it took me a few minutes to figure out what these other kids were doing. Then I realized they were “seeing” the problem like this:

This seems more like 2L + 2W! These students tended to use more of a Guess and Check strategy to find solutions, whereas kids who used the first method were a little more systematic from the start. But it was fun to me to see these two different methods to what is a pretty simple idea. I like when simple things are done different ways.

I wonder how you “saw” the problem when I first described it. And how did your kids tend to see it?

Some ** Frog Farming** links in case you are interested:

- The problem [requires a Math Forum PoW Membership].
- Information about accessing
*Frog Farming*(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!

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 DI 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 SchoolI 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 SchoolIf 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 SchoolWe 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!

by Annie

January
31st,
2012

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.

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