## Mom’s Errands

After Tom’s mother drops Tom off at school, she has to stop at the drug store, the supermarket, and the post office before heading home.

After Tom’s mother drops Tom off at school, she has to stop at the drug store, the supermarket, and the post office before heading home.

Raquel and Esperanza were asked to count Dr. Dolittle’s ostrichs and pushimi-pullyus. Raquel counted 67 heads, while Esperana counted 134 legs.

The Supreme Court of the United States is the highest court in the country. It consists of a Chief Justice and eight Associate Justices.

The Court begins each term on the first Monday in October. Suppose that on that first day, each Justice greets every other Justice by shaking hands exactly once.

A closed rectangular box whose dimensions are 8 feet by 5 feet by 3 feet has 5 feet of water in it.

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.

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. - 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 last Geometry Problem of the Week, *Lotsa Popcorn!*, we presented the World’s Largest Popcorn Ball (as of 2006), and explained that it weighed 3,415 pounds and was 8 feet in diameter. It was claimed to be “almost 50,000 times larger than the normal popcorn balls distributed for retail consumption.”

We went on to ask students to find the diameter of a popcorn ball whose diameter is 50,000 times smaller than this, as well as the diameter of a ball whose volume is 50,000 smaller. We also asked about the weight. We received a number of solid answers to this, including these two:

For #s 1 and 3, I simply took the original value (big ball) and divided by 50,000. For #2, I first found the volume of the big popcorn ball, then divided by 50,000 and used the formula for the volume of a sphere in reverse to find the radius. Then I multiplied that value by 2 to get the diameter.

1. 8/0,000= 0.00016 ft, which did not seem reasonable

2. Volume of original (big ball):

V=(4/3)πr3

V=267.946667 ft3 (roughly)

volume of small:

V= 267.946667/50,000= 0.00535893 ft3

0.00535893=(4/3)πr3

r3=0.00535893/(4/3)π

r=0.108577 ft

d=0.217154 ft or 2.6″ which did seem reasonable

3. 3,415/50,000= 0.0683 lbs, which, unless my perception is off, seems reasonable enough.

1. 0.00192 inches. No, this would not be a realistic size for a normal popcorn ball. 2. 2.605848 inches. This is a realistic size for a normal popcorn ball. 3. 1.0928 ounces. This is not a realistic weight for a normal popcorn ball.

1. Multiply eight feet by twelve inches to find how many inches are in eight feet. Then, divide by 50,000.

2. V=4/3πr3

The radius is half the diameter, which would make it 4 feet.

V=4/3(3.14)(64)

V=4.1866667(64)

V=267.94667 cubic feet

Then divide by 50,000 for 0.00535893 cubic feet.

By using this new volume in the above equation, the radius can be solved for.

0.00535893=4/3(3.14)r3

0.00170667=4/3r3

0.00128=r3

0.108577=r

Since the diameter equals 2r, it is 0.217154 feet. Multiply this by 12 to get 2.605848 inches.

3. 3415 multiplied by 16 equals the number of ounces in 3415 pounds:

54,640 ounces.

This divided by 50,000 equals the answer, 1.0928 ounces.

Obviously we don’t need to give that much direction to students. We just wanted to make sure they would do something reasonable, and we figured we might need to support them in that, especially if they were working on their own without the benefit of a class or partner conversation. What we really did was make all the really interesting math decisions for the students and make it easier for us to read the solutions. We should have just presented them with the facts of the giant popcorn ball, and asked them if the 50,000 times bigger thing was reasonable. What would they compare? I’m thinking back to the So Many Salmon problem we used in Math Fundamentals earlier this year. We presented students with some facts and then asked a question, but let them determine how they would answer that question.

It makes me wonder how many people used the Scenario Only, which did only present the facts, and how many used the whole problem with the three specific questions. What would you have done? What do you think your students, whatever age they are, would have done with just the facts and the question of whether or not the 50,000 times bigger statement was reasonable?

Some ** Lotsa Popcorn!** links in case you are interested:

- The problem [requires a Math Forum PoW Membership].
- Information about accessing
*Lotsa Popcorn!*(and a selection of all our PoWs) for 21 days with a free Math Forum trial account.

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