Lesson 8-6: Areas of Trapezoids (Day 86)

Today on her Daily Epsilon on Math 2021, Rebecca Rapoport writes:

What is the minimum number of unit rods necessarily to make a rigid hexagon?

Well, it's only my first day since completing the long-term assignment, and I'm already going back to the Rapoport calendar, since it's a Geometry question. Well, I'm counting this as Geometry since it does mention a "hexagon."

The tricky part in this question is "rigid" -- what makes a polygon rigid? All triangles are rigid, because given the lengths of the three sides (the "rods"), there is only one possible triangle up to isometry. This is, in essence, the SSS Congruence Theorem. But there is SSSS Theorem for quadrilaterals, much less any SSSSSS Congruence Theorem for hexagons.

So in order for a polygon to be rigid, we must use some of the rods to triangulate it -- that is, divide it into triangles. Since all the rods are the same length ("unit" rods), this suggests that all of the triangles are equilateral. And we know how to make a hexagon out of equilateral triangles -- put six of them together to form a regular hexagon The perimeter of the hexagon uses six rods, and there are six more rods, one joining each vertex to the "center" of the hexagon to form "spokes."

That makes twelve rods -- but notice that today's date is the eleventh, not the twelfth. This suggests that one of the spokes can be removed and yet the hexagon remains rigid -- the rigidity of the other five vertices forces the sixth to remain in place. But if we remove two spokes, the structure isn't rigid.

Though to me, checking the date and then reverse-engineering my work to match the date isn't mathematics -- suppose this had been a question on a test rather than a calendar, then it wouldn't have occurred to me to give 11 as the answer. I must admit that it's just too tempting to start from the date and work backwards, especially when the topic is something that's less familiar. (After all, when's the last time any Geometry teacher taught this?)

But then again, I was reading the Great Women of Mathematics account on Twitter -- each day, the Rapoport question is posted, and readers try to answer them. And one participant -- Amos Duveen -- gives a compelling argument that the answer is nine, not eleven. Here's his trick -- if we remove two of the triangles, then the perimeter is still a hexagon, just not a regular hexagon. It's now a concave hexagon, with three 120-degree, two 60-degree, and one 240-degree reflex angle. And this hexagon requires only three spokes, so the total number of rods is only nine!

Rapoport's error here is that she was clearly thinking only about regular hexagons. Yet Duveen's hexagon is equilateral, but not equiangular. (If all equilateral hexagons were equiangular, then they would already be rigid and we'd only need six rods.) Still, because today's date is the eleventh, the regular hexagon with one spoke removed is the intended solution. But Duveen's answer must be considered correct, since nowhere does it say that the hexagon must be regular.

And indeed, the fact that a hexagon can be divided (or "triangulated") into four triangles is the basis of the proof that the sum of the angles of a hexagon is 720 -- exactly four times that of the triangle.

(One respondent also gives the answer six, with the rods overlapping to form a Star of David. This was likely intended as a joke -- but then again, nowhere does it say that the rods can't overlap....)

So perhaps Rapoport should have printed this question on the ninth instead -- especially since the actual problem she posted on the ninth is also controversial:

The sketch is of equally spaced railroad ties drawn in one-point perspective. (The lengths of three of these ties are 16, 12, and x, and we are asked to find x.)

A problem like this one came up last year on the Rapoport calendar as well -- on February 16th. I didn't mention it at the time, only because that date was a Sunday and hence a non-blogging day. And this problem again comes up on the weekend, but this time I want to address it anyway.

Once again, while perspective is itself mentioned in Geometry (Lesson 1-5 of the U of Chicago text), specific lengths in one-point perspective aren't, So once more, I ended up starting with the date and working backwards. The numbers 16, 12, 9 form a geometric progression -- and last February, the progression was 25, 20, 16, which is also geometric. So the intent is that the lengths of consecutive ties drawn in perspective should form a geometric progression.

But on Twitter, Ben Chia came up with a different answer. He acknowledges where Rapoport gets her answer from -- she assumes that the transformation of perspective (which counts as a projection), while it doesn't preserve congruence, does preserve similarity. And thus, since the ties on the ground form congruent rectangles, they must form similar polygons (in this case trapezoids) in perspective. It then follows from similarity proportions that the tie lengths form a geometric progression.

Then Chia gives his alternative answer -- and based on his knowledge of the properties of perspective, he shows that the ties should form a harmonic progression, not a geometric progression. (A harmonic progression is a sequence whose reciprocals form an arithmetic progression.) Thus since the two given lengths are 16 = 48/3 and 12 = 48/4, the next tie length should be 48/5 = 9.6, not nine. And even if we claimed we were rounding, 9.6 rounds to 10, still not 9.

And in fact, when I was doing research for the February problem, I also stumbled upon the idea that the lengths form a harmonic progression. (Once again, I never posted it because it was the weekend.) So this suggests that Chia's answer of 9.6 is correct. Projections don't necessarily preserve similarity.

So that makes two errors caught on the Rapoport calendar in the first eleven days of the year. This doesn't bode well for the accuracy of the new calendar -- I hope that it will be a long time before we see another error, because I definitely enjoy her calendar!

OK, the next thing I said I'd do once my long-term ended is begin our viewing of the Great Courses DVD that I got for Christmas -- The Mathematics of Games and Puzzles by Prof. Arthur Benjamin. But last week, a mathematical NOVA episode aired on PBS, "Prediction by the Numbers." I didn't describe it on the blog at the time because I wanted to get through the last few days of my long-term. So I waited until today instead.

I don't mind waiting an extra day to begin the Great Courses. After all, Benjamin's course consists of twelve episodes, with four episodes on each disc. Waiting until tomorrow means that I can describe one episode each day and still finish the first disc by the weekend. And besides -- I suspect that some of the material in the NOVA episode and Benjamin's course will overlap anyway.

So let me describe and summarize last week's NOVA episode, "Prediction by the Numbers":

• The future unfolds before our eyes, but is it always beyond our grasp? What was once the province of the gods has now come more clearly into view through mathematics and data.
• At the Orange County Fair, Sir Francis Galton noticed that when many people guessed the weight of an ox, the average weight was closer than anyone's guess.
• Prof. Talithia Williams of Harvey Mudd reproduces this "Wisdom of Crowds" experiment with a jellybean counting contest, and the average was only about 10% off.
• It wasn't until the past 200 years when we used science and mathematics to make predictions -- the branch of math that analyzes data to make predictions is called "statistics."
• Girolamo Cardano, a 16th century Italian, discovered the Law of Large Numbers -- as a coin is flipped more often a coin, the closer the experimental probability gets to 50%
• Casinos take advantage of the Law of Large Numbers to produce a house edge. And because of this Law, shooting percentages in basketball stabilize over a season to reflect true talent.
• In the 17th century, Blaise Pascal and Pierre de Fermat, developed probability theory asked a simple question -- how to divide the pot in a five coin flip game interrupted after three flips.
• Weather forecasting depending on data gathered from balloons to make a probabilistic prediction, which is called "numerical forecasting." These predictions have improved in recent years.
• Sir Ronald A. Fisher devised a method for conducting scientific experiments -- the p-value. Only experiments with p < 0.05 are worth further investigating, or are "statistically significant."
• The temptation to cheat (p-hacking) is enormous, since researchers desire more grant money. We are not sure how often p-hacking actually occurs.
• Pollsters predicted that Hillary Clinton would defeat Donald Trump in the 2016 election, but the polls were wrong. According to Nate Silver, it all came down to sampling and margins of error.
• Baseball has always been a game of numbers, but it's only in the past few decades when analytics grew in importance. One of the first GM's to adopt analytics was Billy Beane, of Moneyball.
• When kids ask Beane what they can do to unlock the power of Moneyball, his reply was always the same -- "Go study it and get an A in math." We produce about 2.5 quintillion bytes daily.
• The Coast Guard's SAROPS system is based on the the theory of Thomas Bayes to find someone who has been lost at sea -- to determine where the Coast Guard should begin searching.
• Our spam filters and search engines also use Bayesian probability. This ultimately leads to the concept of artificial intelligence, or "black box" machine learning.
• While tomorrow will always remain uncertain, mathematics will continue to guide the way through the power of probability and prediction by the numbers.

Stats and probability have taken on more importance under the Common Core, with an additional SP strand taught in both middle school and high school, usually at the end of the year. If students ask us whether this math serves a purpose, we can tell them the information from this NOVA episode -- and in particular, they should take Billy Beane's advice and get an A in math.

As we prepare to make our long-awaited return to Geometry, you might wonder which district the blog calendar is following. As it turns out, I'm following both of my Orange County districts. The last day that differed in the calendars was the teacher PD between the semesters -- my new district closed the Friday before winter break, while my first OC district closed after winter break instead. Both districts are identical from Days 86-180. We'll just ignore my LA County district calendar, even though that's the district whose calendar we've been following until this long-term began.

This is what I wrote last year about today's lesson:

Lesson 8-6 of the U of Chicago text is called "Areas of Trapezoids." In the modern Third Edition, areas of trapezoids appear in Lesson 8-5.

Today is the first lesson of the second semester, and this is the lesson I've been bringing up for a while now. The digit pattern tells us that we should begin with Lesson 8-6. But my long-term position ended on Day 85, thus blocking Lessons 8-3 through 8-5. Our challenge is to cover the missing material from Lessons 8-3 to 8-5, especially triangle area in Lesson 8-5. So let's dive in.

In Lesson 8-3, the Area Postulate tells us that area satisfies four properties. Three of these are Uniqueness, Congruence, and Additivity. The fourth is the rectangle formula, A = lw. But by starting with Lesson 8-6 today, we're essentially postulating the formula not of a rectangle, but of a trapezoid:

A = (1/2)h(b_1 + b_2)

So now all we need to do is derive formulas for the parallelogram, rectangle, and triangle. The area of the parallelogram is easy, since today's Lesson 8-6 already provide for this. Last year I wrote:

• The text uses inclusive definitions, so a parallelogram is a trapezoid. If you're wondering why there's a section for areas of trapezoids but not of parallelograms, this is why. Recall that the most useful fact about a trapezoid that isn't isosceles is its area formula.

Thus we find the area of a parallelogram by using the trapezoid formula. For the parallelogram, both bases are equal, so b_1 = b_2 implies:

A = hb

And now you might notice that a rectangle is a parallelogram, so now we have the rectangle formula as well. Notice that the variables here are A = hb instead of A = lw, but there's nothing stopping us from calling the dimensions of a rectangle "base" and "height" instead of "length" and "width."

This leaves us only with the triangle area -- the subject of Lesson 8-5. I think that there are three different ways to proceed at this point:

• Follow the missing Lesson 8-5 precisely -- branch into three cases depending on whether the triangle is right (half a rectangle), acute, or obtuse.
• Since we have a parallelogram area formula, just take half a parallelogram. This is how some other Geometry texts present the area of a triangle, specifically those that teach a parallelogram formula before a triangle formula.
• Consider a trapezoid with a base of length 0. It's easy to imagine how as the base of a trapezoid approaches zero, the trapezoid's area approaches the triangle's.

The third method is the most interesting to me, since it means that a single formula works for all the shapes that we're teaching today. But I suspect that the second method will be the easiest for the students to understand. The first method, which requires three cases, is a bit much on a day when we're giving the areas of so many other shapes. In any case we obtain the formula:

A = (1/2)hb

For today's lesson, I'll take last year's Lesson 8-6 worksheet. I won't change it, so it means that the vocab term "triangulated polygon" is meaningless, and so the answer to #2 must now change from the old (c) to this year's (b).

(Ironically, triangulated polygons appear in today's Rapoport problem!)

It also includes a "review" question on triangle area, but it's not "review" at all. We might consider this question to be sufficient, but to be safe, let's add an extra worksheet from the old Lesson 8-5 to today's post. Polygon area is one of the more important lessons of the course.

And that's all I have to say today. No, I can't even find a contrived reason to start writing about computer music again. (Even though I have more to say about music.)

So I guess that's all, folks!