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?":
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:
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 [
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.
-- 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.