## Friday, March 6, 2015

### Activity: Area and Tessellations (Day 126)

In some ways, yesterday's lesson was already a semi-activity -- using grids to estimate the area of a circle and therefore a value for pi. Here's a final comment about that Dr. Wu-inspired activity: notice that Wu actually defines pi to be the area of the unit disk -- rather than the circumference of a circle with unit diameter -- because of this area activity. An activity similar to yesterday's, but for perimeter or circumference rather than area, would be awkward.

Nonetheless, today is an activity day. One thing about Chapter 8 is that there are plenty of possible activities to do, since every section from 8-2 to 8-5 in the U of Chicago contains multiple questions in their Exploration sections. I've chosen to set up four activities this week -- the four exploration questions from the two most recent sections, 8-3 and 8-4.

Activity #1 (Section 8-3, Question 26):
Find a rectangular room where you live. Calculate its area a. to the nearest square foot b. to the nearest square meter.

This activity should be straightforward, except for the "where you live" part, since this isn't suitable for a classroom activity. Naturally, I ask the students to find the area of the classroom instead.

Activity #2 (Section 8-3, Question 27):
In 1860, the area of the United States was 3,021,295 square miles. By 1870, the area had become 3,612,299 square miles. What caused such a large change?

This is a good one for students who like history much more than math. This is a tricky one. At first glance, one might see the dates 1860 and 1870 and think "Civil War." Technically speaking, South Carolina did secede in December 1860, but the rest didn't secede until 1861. So if you're thinking about that war, the only Southern state that might be included in the 1870 area but not the 1860 area would be South Carolina, and SC would hardly have one-fifth the area of the whole United States.

A huge hint appears in the U of Chicago text itself, at the beginning of this section:

"An almanac gives the area of the 48 contiguous United States as about 3,026,000 square miles. (The other two states, Alaska and Hawaii, add about 593,000 square miles to the area.)"

And we see that the 1860 area is already close to the current area. But then we notice those important words "of the 48 contiguous United States" (emphasis mine). So we conclude that it's the territories of the last two states, Alaska and Hawaii, that made the difference -- we see that the difference between the 1860 and 1870 areas is right around 600,000 square miles, which is right around the extra area provided by these two states. Finally, we note that Hawaii's tiny islands can't make up the lion's share of that 600,000 square miles -- and besides, the annexation of Hawaii didn't occur until a few more decades after 1870. Therefore, the answer to this Exploration Question is that Alaska -- "Seward's Folly" -- caused the large change in area between 1860 and 1870.

I'm not sure how well our geometry students would fare on this history question. Here in California, American history is taught in the 5th, 8th, and 11th grades, and I'm not sure when Seward's Folly is taught, as it occurred on the cusp between the scopes of the 8th and 11th grade classes. Moreover, notice that most students take geometry in Grades 9-10 -- right between the U.S. history years. But still, this may be an interesting research activity.

Activity #3 (Section 8-4, Question 24)
Multiple choice. When the vertices of a polygon are on a lattice, there is a formula for its area. The formula is known as Pick's Theorem. Use the polygon [given] and test with other polygons, to answer this question. Let P be the number of lattice points on the polygon. Let I be the number of lattice points inside the polygon. Which is the polygon's area (in square units)?
[emphasis the U of Chicago's]

(a) 1/2 P + I - 1
(b) 1/2 P + I
(c) 1/2 P + I + 1
(d) 1/2 (P + I)

Notice that the given polygon in the text happens to be a pentagon. So far we only actually know the area of rectangles. The pentagon can be easily divided into triangles, but we won't know how to find the area of triangles until Section 8-5. Nonetheless, the text includes this question in Section 8-4, perhaps as a preview of Section 8-5.

Of course, students can experiment with rectangles, since these have known areas, in order to figure out which of the four formulas may be correct. In fact, we can quickly determine which of the four must be correct by considering the simplest possible rectangle -- the unit square. We see that the unit square contains four lattice points on the boundary, namely its four vertices (so P = 4), and there are no lattice points in the interior (so I = 0). When we plug in the values P = 4 and I = 0 into each of the four formulas above, we see that only formula (a) gives an area of 1 for the unit square. So we conclude that formula (a), that A = 1/2 P + I - 1, must be correct.

The text states that this result is known as Pick's Theorem. This theorem is named for Georg Alexander Pick, the Austrian mathematician who proved this theorem near the end of the 19th century (that is, right after Alaska and Hawaii became states). According to my favorite math bio page, Pick actually met Albert Einstein. Pick eventually died in the Holocaust, in his eighties:

http://www-history.mcs.st-and.ac.uk/Biographies/Pick.html

Pick's proof of his theorem is a bit complex. It essentially uses induction -- a technique that we've seen used for Polygon-Sum Theorem earlier on this blog. For both the Polygon-Sum Theorem and Pick's Theorem, we cut a polygon into triangles, show that the relevant formula holds for triangles, and conclude that the formula holds for all polygons.

Activity #4 (Section 8-4, Question 25)
[Here] is a pentagon that tessellates, of a type discovered by Marjorie Rice in 1975. Draw enough of a tessellation to show the pattern.

Even though most of the activities listed in Sections 8-3 and 8-4 are about area, this one is about tessellations instead. Then again, tessellations naturally lend themselves to activity, so it would be unthinkable to have an activity day with none based on tessellations.

In Section 8-2, we learned that not every polygon tessellates, but every triangle does. In Question 14 (which I didn't include in the worksheet for that section), we find out that every quadrilateral tessellates, but performing the tessellation is tricky and requires several rotations. But not every pentagon tessellates -- in particular, the regular pentagon doesn't tessellate. According to the text, we don't know exactly which pentagons tessellate, and new discoveries are still being made today.

The tessellating pentagon in this activity was discovered by Marjorie Rice. According to the text, Rice was not a professional mathematician -- indeed, she was a homemaker. The text tells us that Rice is from right here in California. She was still alive in the 1990's, when the text was written -- and it appears that she is still alive today, now in her nineties. Here, in fact, is Rice's website:

The site made an announcement on Rice's 90th birthday that she was no longer updating the site, but a link to some of the tessellations she discovered is still there.

I tried to trace the tessellating pentagon as best I could, but this is the type of figure where if I'm off by just a few degrees, the pentagon won't tessellate. Of course, students can just take a ruler and straightedge and draw the pentagon themselves. Notice that four of the sides of the pentagon are congruent (all labeled by x in the diagram), and one side isn't congruent to the others (y). So students can just draw the four congruent sides of any length x, with the correct angles between them, then join the last vertex to the first.

A few weeks ago, in Chapter 13, we discussed how the LOGO programming language and turtle graphics work. The Rice pentagon is an excellent figure to draw in LOGO, if the students have access to a computer that has LOGO installed. One way to draw the Rice polygon is as follows -- the program takes a single parameter that corresponds to the value x shown on the diagram. This one begins at the top and moves counterclockwise.

TO RICE :X
FORWARD :X
RIGHT 112
FORWARD :X
RIGHT 48
FORWARD :X
RIGHT 100
FORWARD :X
HOME
END

Recall that the angles in a LOGO program are exterior angles, so these are the supplements of the interior angles shown on the diagram. If we begin from a clear screen, we may do RIGHT 110 before beginning the Rice polygon, so that the base of the polygon is horizontal such as it is on the diagram. Finally, the command HOME draws the missing side of the pentagon, since it returns the turtle to its "home" position at the center of the screen.

But using HOME in this situation is not elegant. It forces the turtle to go to the center of the screen, regardless of where we began drawing the polygon. We would want to draw the pentagon anywhere on the screen and at different angles, so that we can tessellate the screen.

What we need is to be able to calculate y, so that we can go FORWARD that many steps. Even if we're drawing this by hand and not by LOGO, we may still wish to find y, rather than follow the steps above and use a ruler to measure y.

One way to calculate y is just to use the vectors of Chapter 14. If we imagine the five sides of the polygon to be vectors, then notice that these five vectors must add up to the zero vector, since a polygon is a closed figure. So all we have to do is add up the first four vectors, and their sum must add to the fifth vector to give the zero vector -- that is, the opposite vector of their sum must be the fifth vector. We already know the magnitude of the first four vectors is x -- and for simplicity, we can simply let x be 1. The direction of one vector can be due east to represent the base, and then the direction of the second vector becomes 48 degrees north of east. That's right -- just as in LOGO, it's the exterior angles that give us the directions of the vectors.

Once we find the fifth vector, its magnitude must be y. Actually, it's the value of y when x = 1 -- but any pentagon that we wish to draw must be similar to the figure for x = 1, thus x and y must be directly proportional.

As you can see, I enjoy this activity because it brings together Chapters 13, 14, and 8. This is ironic since it occurs in Chapter 8 in the U of Chicago text -- a student reading the text in order wouldn't have reached Chapters 13 or 14 yet. But since we changed the order of the text, we can and should take advantage of our reshuffling of the chapters.