Lesson 7-6: Properties of Special Figures (Day 76)

Today I subbed in an Algebra II class. Despite this, I'm not doing "A Day in the Life" today, because two big things happened on Saturday.

One was my 39th birthday. The other was the 80th Putnam competition. And so this is the first of several posts in which I start discussing the Putnam.

Indeed, it's the second since Kent Merryfield passed away. He was the one who normally posted Putnam problems on the Art of Problem Solving Website. Instead, several other posters are discussing the problems from this year's competition -- but instead of posting them Sunday night as Merryfield did, they waited until today at 5PM Pacific Time. Thus, instead of reading the problems Sunday night (which was a non-blogging day for me!) I had to wait until tonight to read and try out the problems -- thus giving me less time to blog today.

It's been a tradition on my blog to post one of this year's Putnam problems. Usually I post it on the Tuesday after the exam, and this year will be no exception, so expect a discussion about a Putnam problem in tomorrow's post.

By the way, while I was looking up info about this year's Putnam, I found this old page on the website of UCLA, my old school:

http://www.math.ucla.edu/~thiele/putnam/history.html

And yes -- my name is listed here. My Putnam score in 2001 was 22 out of 120. This sounds bad, but this placed me among the top 500 test takers in the country. The most common score on the test, after all, is usually zero.

By the way, last year, a UCLA student was good enough for 27th in the nation, and the team overall finished third in the nation, tying our school's best result ever:

https://www.math.ucla.edu/news/ucla-putnam-team-ranks-3-nationwide

Lesson 7-6 of the U of Chicago text is called "Properties of Special Figures." (In the new Third Edition of the U of Chicago text, this is Lesson 7-7. Oh, and the title of the new Lesson 7-7 makes it clear that the "special figures" referred to here are parallelograms.)

I've made many changes to this lesson over the past four years. (The new Third Edition also blows up the old Chapter 13 by moving all of its topics to other chapters. Ironically, Chapter 7 isn't one of them -- Lesson 7-6 of the new edition, on tessellations, appears in Chapter 8 of the old edition, and the other two extra lessons in the new edition don't appear in the old edition at all.)

Anyway, this is what I wrote last year about today's lesson:

The unit test will not be given until just before winter break, leaving us with an extra week to fill. So I asked myself, what else can I fit into this unit?

And so I decided just to post the parallelogram properties after all. This week we cover Lesson 7-6, on the Parallelogram Consequences, and Lesson 7-7, on the Parallelogram Tests.

We've already seen how useful the Parallelogram Consequences really are. The reason that they are delayed until 7-6 in the U of Chicago text is that they are best proved using triangle congruence, but triangle congruence doesn't appear until Chapter 7.

It would probably make more sense to cover parallelograms along with the other quadrilaterals in Chapter 5, but it's too late now. Fortunately, I have this opening in the schedule now to cover both Lessons 7-6 and 7-7.

And today's Lesson 7-6 also includes the Center of a Regular Polygon Theorem, which the U of Chicago text proves using induction. This fits with the Putnam-based lesson that I plan on posting this week.

Oh, and by the way, I found the following page about the Fibonacci sequence and generalizations:

http://mrob.com/pub/seq/linrec2.html

(That's right -- this is my second straight week on the same Putnam problem.) It shows why numbers of the Fibonacci sequence have other such numbers as factors -- it's because the Fibonacci numbers can be written as polynomials that can be factored to get other Fibonacci polynomials.

So how did Dr. Merryfield solve Putnam problems like 2014 B1 or 2015 A2? As it turns out, he uses a proof technique called mathematical induction. Most proofs in a high school geometry course aren't proved used mathematical induction. Indeed, only one proof in the U of Chicago text is proved this way -- and it just happens to be the Center of a Regular Polygon Theorem! Furthermore, Dr. Wu uses induction in his proofs on similar triangles, so this is a powerful proof technique indeed. (On the other hand, Wu simply defines a regular polygon as an equilateral polygon with a circle through its vertices, so he would have no need to prove the following theorem at all.)

I retain this discussion from 4-5 years ago about old Putnam problems in order to demonstrate mathematical induction. In a way, last year's 2018 A6 is sort of like induction in the way we prove certain numbers are rational, but not quite. So the sum of two sexy numbers is sexy, as is their difference, product, and so on.

[2019 update: Again, I won't write about this year's Putnam until tomorrow.]

So here is the proof of the Center of a Regular Polygon Theorem as given by the U of Chicago -- in paragraph form, just as printed in the text (rather than converted to two columns as I usually do):

Proof:

Analyze: Since the theorem is known to be true for regular polygons of 3 and 4 sides, the cases that need to be dealt with have 5 or more sides. What is done is to show that the circle through three consecutive vertices of the regular polygon contains the next vertex. Then that fourth vertex can be used with two others to obtain the fifth, and so on, as many times as needed.

Given: regular polygon ABCD...

Prove: There is a point O equidistant from A, B, C, D, ...

Draw: ABCD...

Write: Let O be the center of the circle containing A, B, and C. Then OA = OB = OC. Since AB = BC by the definition of regular polygon, OABC is a kite with symmetry diagonal OB. Thus ray BO bisects angle ABC. Let x be the common measure of angles ABO and OBC. Since triangle OBC is isosceles, angle OCB must have the same measure as angle OBC, namely x. Now the measure of the angles of the regular polygon are equal to 2x, so angle OCD has measure x also. Then triangles OCB and OCD are congruent by the SAS Congruence Theorem, and so by CPCTC, OC = OD. QED

Now the "Analyze" part of this proof contains the induction. If the first three vertices lie on the circle, then so does the fourth. If the fourth vertex lies on the circle, then so does the fifth. If the fifth vertex lies on the circle, then so does the sixth. If the nth vertex lies on the circle, then so does the (n+1)st. I point out that this is induction -- from n to n+1.

Every induction proof begins with an initial step, or "base case." In this proof, the base case is that the first three points lie on a circle. This is true because any three noncollinear points lie on a circle -- mentioned in Section 4-5 of the U of Chicago. The induction step allows us to prove that one more point at a time is on the circle, until all of the vertices of the regular polygon are on the circle.

We have seen how powerful a proof by induction can be. We have proved the Center of a Regular Polygon Theorem. But now we wish to prove another related theorem -- one that is specifically mentioned in the Common Core Standards:

CCSS.Math.Content.HSG.CO.A.3
Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself.

What we wish to derive from the Center of a Regular Polygon Theorem is that we can rotate this polygon a certain number of degrees -- about that aforementioned center, of course -- or reflect it over any angle bisector or perpendicular bisector. But the U of Chicago text, unfortunately, doesn't give us a Regular Polygon Symmetry Theorem or anything like that.

I'm of two minds on this issue. One way would be to take this theorem and use it to prove that when rotating about O, the image of one of those isosceles triangles with vertex O and base one side of the polygon is another such triangle. The other way is to do Dr. Wu's trick -- he defines regular polygon so that it's vertices are already on the circle. Then we can perform rotations on the entire circle. (Rotations are easier to see, but it's preferable to do reflections because a rotation is the composite of two reflections.) Notice that the number of degrees of the rotation depends on the number of sides. In particular, for a regular n-gon we must rotate it 360/n degrees, or any multiple thereof.

The modern Third Edition of the text actually mentions rotation symmetry. This section tells us that a parallelogram has 2-fold rotation symmetry, and the statements about regular polygons actually appear in Lesson 6-8 of the new edition.