## Two Heads are Better than One

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

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.

The steel girders being used to construct a building are 27 feet long. One of the corridors the girders are to be carried down has a 90° turn at the end. The width of the hall before the turn is 8 feet.

A plane traveling 400 mph is rising at an angle of 30 degrees. A second plane, traveling 300 mph, is rising at an angle of 40 degrees.

Today I started to plan my garden. The rectangular garden where I am going to plant is three feet longer than it is wide.

One third of the garden area will be sunflowers.

Another quarter will be planted with brow-eyed susans.

Only a fifth of my garden will contain columbine.

A sixth of my garden will be filled with trillium.

The rest will be foxgloves.

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!

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