Lesson 3-6: Constructing Perpendiculars (Day 36)

This is what Theoni Pappas writes on page 278 of her Magic of Mathematics:

Step 1) label three points on a sheet of paper as shown, (1, 2); (5, 6); (3, 4).

This is the first page of a new section, "The Games Mathematicians Play." And the first game that Pappas describes here is called "The Chaos Game."

In the lone picture on this page, Pappas shows us that (1, 2), (3, 4), and (5, 6) are not intended to be the Cartesian coordinates of points. Indeed, if these were xy-coordinates, they would be collinear as they all satisfy the equation y = x + 1. On the contrary, she shows us that these are supposed to be the vertices of an equilateral triangle.

Why would Pappas do this? Well, let's read the rest of the steps to find out why:

2) Pick a starting point at random. Roll a die.
3) Rule: take 1/2 the distance between the starting point and the point which has the number on the die that was just rolled.
4) Continue step three indefinitely.

Pappas tells us that randomness will conform to order -- the resulting figure will turn out to follow a surprising pattern. Indeed, she gives other examples of order from chaos:

"Similarly, in 1904 R. Chartres found out that the...."

Oops -- the rest of that sentence is on page 279. Well, I might as well spoil the surprise, because I actually wrote about this a few weeks ago:

"...probability of two numbers, written at random, being relatively prime is 6/pi^2."

That's right -- we proved this two weeks ago in our reading of Ogilvy. Now you can see why we spend so much time reading Ogilvy's book, as the ideas of number theory keep coming up again.

Now it's time for yesterday's solutions. Let's start with "Can you make a Latin square?":

601345
526134
250463
342510
063251
435026

This time, the leftmost and rightmost columns have all vertical dominoes. The second and third columns from the left have all horizontal dominoes. The fourth and fifth columns are arranged just like the initial setup -- horizontals in the middle two rows, verticals in the rest of these columns.

Now as for the angle bisection in "Is It Possible?" -- hey, today's U of Chicago lesson is actually on constructions, including bisectors! Since this Pappas puzzle fits today's lesson like a glove, let's discuss the lesson first and then we'll solve the Pappas puzzle.

Lesson 3-6 of the U of Chicago text is "Constructing Perpendiculars." (It appears as Lesson 3-9 in the modern edition of the text.)

Again, I had to go back three years to get the rest of Chapter 3, so we might as well finish it. This is what I wrote three years ago on today's lesson:

Section 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 Section 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 (Section 4-4), 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: Questions 16 and 17 from Section 4-7 of the text show us how to perform an angle "bisectomy."

Okay, that's enough from three years ago. Since we haven't reached Lesson 4-7 yet, let's just solve the angle bisection in Pappas instead. This isn't a true construction since we're using a ruler, rather than an unmarked straightedge and compass.

This is what Pappas writes to describe the bisection of an angle whose vertex is A:

"Locate the 4 points on the sides of the angle as shown [Points E and C are on different sides of the angle such that AE = AC. Points B and D are on rays AE and AC respectively such that EB = CD.] and draw in the two segments [BC and DE]. The segments' point of intersection and the vertex of the angle determine the line that bisects the angle."

Here is her informal proof that the angle A has been bisected into two smaller angles, 1 and 2:

-- Triangles ABC and ADE are congruent by SAS.
-- Angles B and D are congruent by CPCTC.
-- Triangles BEP and DCP are congruent by AAS.
-- BP and DP are congruent by CPCTC.
-- Triangles BPA and DPA are congruent by SSS.
-- Angles 1 and 2 are congruent by CPCTC. QED

Notice that we could have used compasses to make the segments congruent, and this would become a true construction. Unlike the traditional angle bisector construction, this one can be completed using the "collapsible compass" of Euclid. Just like the angle copy construction, we must wait until we have the triangle congruence theorems before we can actually prove this. It is valid in neutral geometry, but not spherical geometry due to the presence of AAS -- but instead, we can convert it into a spherical proof by replacing AAS with AAA.

Two years ago, I did create a few worksheets for Lesson 3-6. These worksheets can still be used today, except that students can't do Construction III yet since students haven't quite learned about reflections yet. Construction IV can be completed using Two Perpendiculars.