A few weeks ago, I mentioned the math teacher Lisa Bejarano, who had posted something called "Euclid: The Game" in one of her recent posts. And when I saw that part of the game reminded me of ancient geometer's Proposition 1 from Lesson 4-4, I couldn't resist checking the game out.

First, here's a link to Euclid: The Game:

http://euclidthegame.com/

Apparently, this is a one-player game. The goal is, on each level, to construct the figure in the diagram at the top of each page. The possible moves are the same as those allowed in classical Greek construction -- drawing an arbitrary point, drawing a point at an intersection, drawing a segment given two endpoints, drawing a ray given the endpoint and another point, and drawing a circle given the center and a point on the circle.

Now Level 1 is indeed Euclid's first proposition -- to draw an equilateral triangle given a side. This one, despite being Level 1, may be tough for students seeing this for the first time -- but of course, our students who remember yesterday's Lesson 4-4 should have no trouble with this one. Notice that according to Kasper Peulen, the creator, this game is powered by Geogebra -- and we were just talking about John Golden and his Geogebra lessons this week. Yes, I'm definitely going to keep going back to Bejarano, Golden, and other teachers when looking for good geometry activities.

Level 2 requires students to construct midpoints. The usual way to perform this construction is to construct the perpendicular bisector -- it intersects the original segment at its midpoint. For our students, this will be a preview of next week's Lesson 4-5 on perpendicular bisectors.

Level 3 requires students to construct angle bisectors. As we've already seen here on the blog, angle bisectors appear on the Common Core tests, yet are given short shrift in the U of Chicago text. The construction is buried in a Question in Lesson 4-7. Here's how to bisect Angle

*AOB*:

Step 1. Circle

*O*containing

*A*

Step 2. Circle

*O*intersects Ray

*OB*at

*C*.

Step 3. Subroutine: Line

*PQ*, the perpendicular bisector of

~~AC~~

*As it turns out, the Euclid game has an equivalent of a "subroutine" -- like many computer and video games, passing a level unlocks a new "tool." In Level 2, I had already unlocked the midpoint tool. So I decided to follow the U of Chicago suggestion -- I drew a circle*

*A*(to label points of intersection

*B*and

*C*), found the midpoint

*D*of

*, and then drew Ray*~~BC~~

*AD*. I passed the level with a minimum number of moves, three.

Level 4 requires students to find the perpendicular to a line through a point on the line. In the U of Chicago text, this is Example 2 of Lesson 3-6. This time, following the U of Chicago construction doesn't give me the minimum number of moves -- I needed four, but the minimum is three.

Level 5 requires students to find the perpendicular to a line through a point

*not*on the line. It is the line given in this week's Uniqueness of Perpendiculars Theorem.

Level 6 requires students to find the

*parallel*to a line through a point not on the line. Recall that this is actually possible in neutral geometry (as Playfair is all about

*uniqueness*). Last year, I mentioned how to do this using only perpendiculars, and doing so only requires the minimum two moves.

I decided to stop at Level 6, because Level 7 is no longer valid in neutral geometry. In particular, given segment

*C*, students are to find

*D*such that

*D*such that

*ABDC*is a parallelogram -- so that I could use the parallel tool unlocked in Level 6. Even though the Parallel Tests are neutral, the Parallel Consequences aren't. Therefore Level 6 is the last level that I'd want the students to reach.

Of course, not every classroom has access to a computer -- then again, Euclid obviously didn't have a computer in ancient Greece either. So I decided to create worksheets for the first six levels of Euclid: the Game, and students will have to solve them the way that Euclid would have.

Meanwhile, I created worksheets for the proof of reflection over the axes. This is all in addition to the original worksheet for today, the Centauri challenge. This is what I wrote about it last year:

The following activity is another one from Michael Serra's text -- it appears in his Chapter 15, since this one often goes with two-column proofs (and proofs appear in his book late). In this activity, strings consisting of the letters P, Q, R, and S are converted into others using four rules (that end up being our postulates):

Rule 1. Any two adjacent letters in a string can change places with each other. (PQ=>QP)

Rule 2. If a string ends in the same two letters, then you may substitute a Q for those two letters. (RSS=>RQ)

Rule 3. If a string begins in the same two letters, then you may add an S in front of those two letters. (PPR=>SPPR)

Rule 4. If a string of letters starts and finishes with the same letter, then you may substitute an R for all the letters between the first and last letters. (PQRSP=>PRP).

Then the text gives the following theorem, PQQRSS=>QRQ. (Notice that I chose to write "=>" where the text writes ">>" since, after all, we've already used the former to denote the hypothesis and conclusion of a conditional.)

Proof:

Statements Reasons

1. PQQRSS 1. Given

2. PQQRQ 2. By Rule 2

3. QPQRQ 3. By Rule 1

4. QRQ 4. By Rule 4

So for the students, this is a puzzle which gets them thinking about the logical structure of proofs without having to think about geometry.

The text calls this the "Centauri challenge," which I assume refers to Alpha Centauri, the closest star system to the sun. Notice that many of the Cooperative Problem Solving challenges in Serra are said to take place in a futuristic lunar colony. For this one, the inhabitants of this colony are trying to communicate with aliens from (Alpha) Centauri, but apparently, the Centaurian alphabet consists of only four letters.

My worksheet contains all of Challenge 1, then adds Challenge 2 as a Bonus. In case you're curious, here are my answers to Challenge 2.

1. Can you produce a string of five or more letters that cannot be reduced to RQ?

My answer is that I can't -- but now I must prove it. Here is my proof, in paragraph form:

Proof:

Our string has five letters, but there are only four letters available. So one of those letters must appear at least twice! (This is called the Pigeonhole Principle.) Let's call the letter that appears twice X. (I know, it's actually P, Q, R, or S, not X, but here I'm using X as a

*variable*to stand for one of the letters P, Q, R, or S, since I want this proof to be as general as possible.)

Using Rule 1, we take the first appearance of X and change it with the letter on its left. Now take that X and change it with the new letter on its left. Keep doing this until the X is the first letter. Now take the last appearance of X and change it with the letter on its left. Keep doing this until this other X is the last letter.

Now our string begins with X and ends with X. So by Rule 4, it becomes XRX. Using Rule 1, we can change the first X and the R to obtain RXX. Finally, this string ends in the same two letters (whatever letter the X stands for), so by Rule 2, it becomes RQ. QED

After doing this, the second question becomes obvious.

2. One of the rules of Centauri can be removed without losing any of the first five theorems proved in Challenge 1. Which of the rules can be removed?

Look at which rules have been used in the post so far, and notice which one's missing. That's the rule that can be removed. I like the similarity between determining which theorems are provable with or without a certain rule, and finding out which theorems in geometry require a Parallel Postulate.

Here is the activity:

## No comments:

## Post a Comment