Lesson 3-6: Constructing Perpendiculars (Day 34)

Part VI of Benoit B. Mandelbrot's The Fractal Geometry of Nature is on Self-Mapping Fractals, and consists of Chapters 18 through 20. In this section, Mandelbrot describes even more methods of obtaining fractals besides self-similarity.

In Chapter 18, Mandelbrot begins by describing the transformation known as inversion -- in other words, a circle reflection. I mentioned circle reflections two weeks ago here on the blog as an example of a non-Common Core transformation that the Geogebra software can perform. Well, some of the fractals Mandelbrot mentions are self-inverse -- that is, the image of itself under a circle reflection is itself. His examples of self-inverse fractals include Apollonian nets -- an infinite set of circles, each of which is tangent to some of the others.

Here Mandelbrot generalizes circle reflection to a more general transformation, which he calls a homography, or Mobius transformation. He defines a homography as the product (composite) of an inversion, a symmetry with respect to a line (a Common Core reflection), and a rotation. But as we already know (and will teach next week), a rotation is simply the composite of two reflections in intersecting lines, so mentioning "rotation" here is redundant. As it turns out, not only is every isometry a homography, but so is every similarity transformation! (Don't forget that the only transformations that appear in Common Core are similarity transformations.) This is because a dilation is -- as I mentioned two weeks ago -- the composite of two circle reflections.

In Chapter 19, Mandelbrot mentions another transformation that one can use to produce even more fractals -- namely, squaring, or performing the function z -> z^2 - mu for some complex mu. He calls a fractal that is invariant under this transformation self-squared. One self-squared fractal is so famous that it is simply known as the Mandelbrot set.

Here are links to the Apollonian gasket and the Mandelbrot set:

http://mathworld.wolfram.com/ApollonianGasket.html
http://mathworld.wolfram.com/MandelbrotSet.html

Last night was the first presidential debate for the Democrats. As I more or less expected, Common Core didn't come up during the debate. Even though, as I mentioned earlier, there exist Democrats who oppose Common Core, opposition to the Core is stronger among Republicans, since the Core was adopted under a Democratic administration.

We move on to Lesson 3-6, which is on constructions. Actually, let's go back to what I wrote about this constructions last year first, and then let me add this year's commentary:

Lesson 3-6 of the U of Chicago text deals with constructions. So, we're finally here. The students will need a straightedge and compass to complete this lesson.

Here's a good point to ask ourselves, which constructions do we want to include here? The text itself focuses on the constructions involving perpendicular lines. Well, let's check the Common Core Standards, our ultimate source for what to include:

CCSS.MATH.CONTENT.HSG.CO.D.12
Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line.

CCSS.MATH.CONTENT.HSG.CO.D.13
Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle.

But let's also go back to what David Joyce writes about constructions:

The book [Prentice-Hall 1998 -- dw] does not properly treat constructions. Constructions can be either postulates or theorems, depending on whether they're assumed or proved. For instance, postulate 1-1 above is actually a construction. On pages 40 through 42 four constructions are given: 1) to cut a line segment equal to a given line segment, 2) to construct an angle equal to a given angle, 3) to construct a perpendicular bisector of a line segment, and 4) to bisect an angle. Later in the book, these constructions are used to prove theorems, yet they are not proved here, nor are they proved later in the book. There is no indication whether they are to be taken as postulates (they should not, since they can be proved), or as theorems. At the very least, it should be stated that they are theorems which will be proved later.

David Joyce, after all, emphasizes that at the very least, constructions should be proved. He writes here that they can be proved later -- but of course, he prefers that theorems not be stated until they can be proved.

So which of the theorems in the Common Core list can be proved so far? Let's look back at that list one by one:

Copying a segment: This should be trivial to construct and prove. The student simply uses the straightedge to draw a line, marks a point O on it, and opens up the compass to the length of the given line segment AB to mark the second point P. The proof that these segments AB and OP have the same length simply follows from the definition of straightedge and compass. It's a bit surprising that the U of Chicago text doesn't begin with this as the first construction, as this one should be easy for the students.

Copying an angle: This is the one that we can't prove yet. The usual construction requires SSS to prove. This is one reason why Joyce would prefer that Chapter 8 of his text occur before these constructions in his Chapter 1.

Bisecting a segment; constructing perpendicular lines, including the perpendicular bisector of a line segment: This is the focus of Lesson 3-6 of the U of Chicago text. Notice that as soon as we've constructed the perpendicular bisector, we've already done the other two constructions (bisecting the segment and drawing its perpendicular). And so, as soon as we prove the perpendicular bisector construction, we are done.

Given: Circle A contains B, Circle B contains A, Circles A and B intersect at C and D.

Prove: Line CD is the perpendicular bisector of AB

Proof (in paragraph form -- can be converted to two columns later):

Just as in the proof of Euclid's first theorem (Lesson 4-4, two weeks ago), since both B and C lie on circle A, AB = AC by the definition of circle, and since both A and C lie on circle B, AB = BC. Then by the Transitive Property of Equality, AC = BC -- that is, C is equidistant from A and B. So, by the Converse of the Perpendicular Bisector Theorem,C lies on the perpendicular bisector of AB. In the same way, we can prove that D also lies on the perpendicular bisector of AB. And through the two points C and D there is exactly one line -- and that line is the perpendicular bisector of AB. QED

In many texts, it's pointed out that the compass opening for the two circles need not be exactly the same as AB. All that's necessary is for the opening to be greater than half of AB -- that guarantees that the two circles intersect in two points. 

The text states that the midpoint of AB has been constructed as a "bonus" -- so we've bisected the segment, as requested. All we need now is to construct perpendicular lines -- and that's exactly what the text does in the next example, construct a perpendicular to line AP through point P on the line, using our perpendicular bisector algorithm as a subroutine.

Bisecting an angle: This is an interesting one. As I mentioned last week, Questions 16 and 17 from Lesson 4-7 of the text show us how to perform an angle "bisectomy." Notice that this construction is slightly different from the one usually given in texts, since the usual construction requires SSS to prove while this one depends only on reflections:

Given: Circle O contains A, Circle O intersects ray OB at C, Line PQ is the perpendicular bisector of AC

Prove: Ray OP is the angle bisector of Angle AOB

Proof:

Since both A and C lie on circle O, AO = CO by the definition of circle. So, by the Converse of the Perpendicular Bisector Theorem, O lies on the perpendicular bisector of AC. Now let line OP be a reflecting line. The image of A is C, and points O and P, both lying on the mirror, are each their own image. So the image of angle AOP is angle COP. Since reflections preserve angle measure, AOP and COP have the same measure. Therefore, by the definition of angle bisector, ray OP is the angle bisector of AOB. QED

Since we are jumping around the book right now, we can perform a few more constructions that we've skipped over. In Lesson 4-1, we construct the reflection image of point B over the line m:

Given: Circle B intersects m in Q and S, Circles Q and S contain B and B'

Prove: B' is the reflection image of B over m

Proof:

Once again, using the definition of circle and the Transitive Property of Equality, we have both BQ = B'Q and BS = B'S, so once again, by the Converse of the Perpendicular Bisector Theorem, both Q and S lie on the perpendicular bisector of BB'. Since two points determine a line, the line containing Q and S -- line m -- is the perpendicular bisector of BB'. So by the definition of reflection, B' is the mirror image of B over m. QED

The book states that there is another "bonus" -- line BB' is the perpendicular to m through B. So we have another construction and the existence of a line perpendicular to a line through a point not on the given line. (This line is unique by the Uniqueness of Perpendiculars Theorem.) But Common Core asks for the similar construction:

Constructing a line parallel to a given line through a point not on the line: Believe it or not, we can complete this construction right now. Usually, the construction involves copying in angle in a way so that the angle and its copy are corresponding or alternate interior angles, so that the lines are parallel. We have neither given the angle copy construction nor the parallel test for corresponding angles. But we do have the theorem from yesterday -- the Two Perpendiculars Theorem.

So given a line l and a point P not on the line l, we construct the line parallel tol through point P as follows:

Step 1. Subroutine m, a line perpendicular to l.

Step 2. Subroutine k, a line perpendicular to m through P.

then by the Two Perpendiculars Theorem, k and l are parallel. Notice that line m need not contain point P. If it does, then Step 1 is a "construct perpendicular through point not on line" step and Step 2 is "construct perpendicular through point on line." Or one could choose a point Q on l, and then do the "construct perpendicular through point on line" step first instead of second.

I was not able to create a good worksheet for constructions. So I leave you on your own to find a good worksheet.

.. and now here's what has changed since last year. First of all, I do now have a good constructions worksheet -- the Euclid: The Game worksheet from two weeks ago. I've also since created a worksheet based on the constructions mentioned in today's blog post.

Comparing today's worksheet to Euclid: The Game, we notice that Construction IV: Parallel Lines, is identical to Level 6 of Euclid: The Game. Notice that, as I said above, Construction II: Perpendicular Bisector, allows you to find the midpoint as a "bonus," while Level 2 of Euclid: The Game, actually asks for the midpoint. Construction III requires one to reflect a point, but as far as I know, reflections don't appear in Euclid: The Game at all.

Now here's the thing about Construction I, which is merely copying a segment. This construction sounds trivial with a compass -- and recall how I wrote earlier that this should be considered the simplest construction. Well, in Euclid: The Game, this construction isn't so simple. The problem is that to the ancient Greeks, the compass to be used in construction only allowed one to construct a circle with given center and radius (i.e., Euclid's Third Postulate). It didn't allow one to transfer distance -- this is often called a "collapsible compass" because as soon as the compass is lifted from the paper, it collapses, and so there's no way to recover the distance.

Now as it turns out, we can convert a "collapsible compass" into a regular compass. In fact, this is exactly what Levels 7 through 9 of Euclid: The Game are all about. After completing Level 9 of the game, we unlock a new tool -- the "compass" tool. Once this is unlocked, we can now perform tasks similar to Construction I, since using the compass tool is an abbreviation for performing the more complicated construction using the collapsible compass.

One interesting way of combining the worksheet for today and Euclid: The Game is by having the students play the game first -- including guessing how to solve the higher-level problems -- and then going back to see whether they are right when we cover constructions for real when they reach today's lesson.

Many schools around the country today administered the PSAT exam to sophomores and juniors. So today is a good day for a filler lesson. Last year, the day after I posted the worksheet for Lesson 3-6, I provided an activity where students construct a tic-tac-toe board. In some ways, an activity is a great lesson to post on a day when students are distracted by the PSAT.

Notice that on the day I posted tic-tac-toe last year, I mentioned the collapsible compass. That day, I gave a link to David Joyce's Euclid site, where he describes how the ancient Greek sage was able to use his collapsible compass -- it's his Proposition II. This is the same as Levels 7 through 9:

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI2.html

If the students view the tic-tac-toe board as just a higher level of Euclid: The Game, then might enjoy it, especially if they get to play Tic Tac Toe: The Game on the board afterward. But if constructions are seen as work, then they won't want to do it while other students are taking the PSAT, or after having just finished the PSAT themselves. So teachers should be careful with today's lesson.