Here is today's Pappas question of the day:

*How many innings did Team Mexico pitch in this year's World Baseball Classic?*

The answer is 17 -- and of course, today's date is the seventeenth.

Yes, I plan to address the controversy regarding the World Baseball Classic in this post -- after all, Mexico was eliminated from the tourney due to a *mathematical formula*, and this is a math blog. But first, I wish to write about the status of my blog.

It is still the Big March, and times are so tough that I've lost the will to write about my class. And so I'm hereby **dropping out of the MTBoS**, for now. In particular, I am no longer a part of Tina Cardone's "Day in the Life" challenge -- yet another MTBoS challenge I fail to complete. Even though tomorrow is the 18th, my monthly posting date, I won't be submitting any post.

By the way, many of the other "Day in the Life" posters have already dropped out (and that ironically includes Cardone herself). The few remaining participants only make one post a month, so of course it's much easier for them. I thought that I could handle three posts a week during the school year, but in the end I couldn't.

My plan now is for this blog to return to its roots. The blog name is "Common Core Geometry," and so I'm returning to posting Geometry lessons from the U of Chicago text. Before I was hired as a teacher, my plan had been to convert the day count into the lesson number to cover -- so since today is Day 121 on the blog calendar, we'll start with Lesson 12-1. The good thing about this is that I can cut and paste most of these lessons from the first two years of this blog, so it doesn't require as much effort on my part. Ironically, Lesson 12-1 is one of the few lessons I didn't post on the blog either last year or the year before -- more on that later.

Before we begin though, let me wrap up a few loose ends. I will still write about my class from time to time on the blog. In particular, you may be wondering about Pi Day. Well, I'd wanted to surprise my students by getting them a pizza on 3/14 at 1:59, as is traditional since pi is 3.14159.... The problem is that, as you can tell by the day count, we're at the start of the third trimester. Therefore, this week is the second Parent Conferences Week, with students dismissed at 1:15 everyday. As much as students like pizza, they wouldn't want to stay 44 minutes after dismissal to get it.

So instead, I served pizza to the latest class that met on Pi Day. With the schedule mixed up to conferences, that class turned out to be sixth grade. (See my November 17th post, another Parent Conferences day, to see how the schedule works on days like this.) I would have preferred seventh grade to get the pizza since that's the grade when they learn about pi, but oh well.

I've written the relationship between Pi Day and the ends of trimesters/quarters before (for example, in my November 9th post). At schools that start after Labor Day, the end of the second trimester often falls close to Pi Day. But at Early Start schools such as mine, the second trimester ought to end at least a week before Pi Day -- in fact, it's the *third quarter*, not the *second trimester*, that should end closer to the math holiday. But with my school's extra holidays (most notably the longer winter break), the second trimester is extended towards Pi Day. So with Parent Conferences, the school will usually be dismissed before 1:59 on Pi Day.

Okay, let's get to that World Baseball Classic. Its structure is similar to that of the World Cup, the international soccer tournament. All the teams are divided into groups of four teams, and in the first round, each team plays each of the other three teams in the group. This is called *round robin*. The top two teams in each group advance to the second round.

As it turns out, there are four possible outcomes for the standings after all the games in a group have been played. The most straightforward is for the four records to be 3-0, 2-1, 1-2, and 0-3. Then the two teams that advance are obviously the 3-0 and 2-1 teams. The second possibility is for there to be two 2-1 teams and two 1-2 teams. This is also easy -- the two 2-1 teams advance.

In this year's classic, three of the four groups followed the 3-0, 2-1, 1-2, 0-3 patterns, so there were clear first and second place finishers in those groups. But the two remaining possibilities involve three-way ties -- either three 2-1 teams and one winless team, or one undefeated team and three teams with 1-2 records.

This last case occurred with the group containing Italy, Mexico, Puerto Rico, and Venezuela. As it happened, Puerto Rico was the 3-0 team, so they obviously advanced. The problem is how to determine the second team to advance. The rules state that there should be a tiebreaker game -- but only two teams can play in that game. So the question is, which two teams among Italy, Mexico, and Venezuela should play in the tiebreaker?

It seems fair that the tiebreaker should be determined by *runs* -- since after all, the objective of baseball is to score more runs than the other team. The Classic actually uses a *defensive* tiebreaker rule -- it's the two teams that allowed the fewest number of runs. Since Puerto Rico isn't involved in the tiebreaker, only the runs Italy, Mexico, and Venezuela allow to the other two tied teams count. As it turns out, Mexico allowed 19 runs, Italy 20 runs, and Venezuela 21 runs in those games. Yet Mexico wasn't allowed to play in the tiebreaker game.

That's because there's one more component to the tiebreaker rule -- *innings*. It's not which team gives up the fewest runs, but the fewest runs *per inning* pitched. In a way, this makes sense in light of the existence of *mercy rules* -- if a team is leading by ten runs after seven innings, the game ends. So a team that loses 10-0 after seven is penalized more than a team that loses 10-0 after nine. The innings rule also rewards teams that play a scoreless tie through nine and lose 1-0 in extra innings.

The innings rule ends up hurting Mexico. This is because Italy and Venezuela played a 10-inning game, so each team is credited with 19 innings (ten against each other and nine against Mexico.) But Mexico is only credited with 17 innings. Their game against Italy is only considered to last eight innings, because the Italians won it on a walk-off in the ninth with nobody out. Had there been one out, it would be considered 8 1/3 innings, and with two outs, it would be 8 2/3 innings.

And so we now calculate the defensive runs per inning:

Italy: 21/19 = 1.05 runs per inning

Venezuela: 20/19 = 1.11 runs per inning

Mexico: 19/17 = 1.12 runs per inning

Therefore Venezuela and Italy played in the tiebreaker, and Mexico was left out.

Here's the question -- suppose you are on Team Mexico, and you are about to play Venezuela in the final game of the group. You already know the scores of the previous five games, so you know that Puerto Rico is 3-0, and already have two losses, so your only hope is the tiebreaker. So what exactly must you do in the game against Venezuela (besides beat them of course) in order to play in the tiebreaker game?

This is an interesting algebra problem. We already know the scores of the other games, so in particular Italy beat Mexico 10-9 (with none out in ninth), and Venezuela beat Italy 11-10. So let's set up some variables -- in the final game, Mexico scores *m* runs and Venezuela scores *v* runs. We now calculate the number of runs per inning:

Italy: 20/19 = 1.05 runs per inning

Venezuela: (*m* + 10)/19 runs per inning

Mexico: (*v* + 10)/17 runs per inning

Since two teams advance to the tiebreaker, Mexico only needs to have a better average than just one of the other teams, either Italy or Venezuela. Let's look at Italy first. In order for Mexico to play on, we need Mexico's run-to-inning ratio to be less than Italy's. So we write an inequality:

(*v* + 10)/17 < 20/19

If this were an equation, an Algebra I student could solve this by cross-multiplying. This is awkward when it's an inequality, though. So instead, we can clear fractions by multiplying by 17(19):

19(*v* + 10) < 17(20)

19*v* + 190 < 340

19*v* < 150

*v* < 7.89

Notice that we *don't* get to round 7.89 up to eight. This is an inequality that says that *v* must be less than 7.89, and 8 is not less than 7.89. So the inequality states that if Mexico beats Venezuela and holds them to seven or fewer runs, Mexico is in the tiebreaker game.

As soon as Venezuela scores its eighth run, the inequality no longer holds. Mexico can still advance, though, if they beat Venezuela by a large enough margin. To figure this out, we set up another inequality, this time to show that Venezuela's ratio is greater than Mexico's:

(*m* + 10)/19 > (*v* + 10)/17

17(*m* + 10) > 19(*v* + 10)

17*m* + 170 > 19*v* + 190

17*m* > 19*v* + 20

*m* > 1.12*v* + 1.18

This inequality is tricky because the coefficient of *v* isn't 1 -- it's not as simple as say *m* > *v* + 1 (which would state merely that Mexico would have to win by two). Since this inequality is irrelevant when *v* is less than 8 (as Mexico would already be in the tiebreaker game with a victory), we start out by plugging in *v* = 8:

*m* > 1.12(8) + 1.18

*m* > 10.14

Again *m* must be *greater* than 10.14, so Mexico needs at least 11 runs. In other words, if Venezuela scores 8, Mexico must win by *three* runs, not two. If we plug in *v* = 9, we obtain:

*m* > 1.12(9) + 1.18

*m* > 11.26

So again Mexico would need a three-run victory. Plugging in 10, 11, and so on for *v* continues to show that Mexico needs to be up by three. Once we reach 16, it changes:

*m* > 1.12(16) + 1.18

*m* > 19.1

Then Mexico would need to win by four -- but this is baseball. Scores like 20-16 are uncommon, so for all intents and purposes, only a three-run victory is needed. We summarize this as follows:

*In order to advance to the tiebreaker game, Mexico must beat Venezuela and:*

*-- If Venezuela scores seven runs or fewer, then any margin of victory is sufficient.*

*-- If Venezuela scores between 8 and 15 runs, then Mexico must win by at least three runs.*

All of this presupposes a nine-inning game, by the way. If the game extends into extra innings, Mexico benefits because they can put up zeros, but then Venezuela benefits as well. Then again, an extra inning would help Mexico have a better ratio than Italy at least -- for example, if the game lasts ten innings, then we can replace "seven" in the above with "eight," so that a 9-8 victory in ten innings (but not nine) is enough to make the tiebreaker.

In the actual game, Mexico held an 11-6 lead at the seventh inning stretch. But then Venezuela scored three in the bottom of the seventh, When the eighth run crossed the plate, we moved from the first case above to the second, but Mexico was still in a good position with a three-run lead. As soon as the ninth run scored, Mexico needed to score another run to advance.

In the bottom of the ninth, team captain Adrian Gonzalez believed that a two-run victory was sufficient to advance. He stated that if Venezuela scored a run in the ninth, the team would have intentionally walked or balked in another run to tie the score, then hope to win by two in extras. I wonder what Gonzalez would have said if he knew that the margin had to be three -- walking and balking in *two* runs makes the game into a farce -- and it's hard to win by three in extras anyway. (In the 11th inning Mexico would start with two runners on base, so a homer would be enough for that three-run lead, but then Venezuela would start the bottom half with runners on base too.) The best way for Mexico to advance was not to give up more than seven runs in the first place.

All of this reminds me of another sport that has a round robin format -- Quidditch. In J.K. Rowling's *Harry Potter* series, the four teams at Hogwarts play a round robin tournament. In three of the books, Harry's Gryffindor team wins the Quidditch Cup -- but oddly, it's never as simple as just going 3-0 in the three matches. Instead, Gryffindor is one of two 2-1 teams (with the lone loss always to, of all teams, *Hufflepuff*), and a tiebreaker of total points in the three matches is used.

A few Potter fans have criticized this tiebreaker system, and have written fan fiction in which teams take advantage of the system. Here is such a story:

The author, Crys, has Gryffindor and Hufflepuff agree before the game that each team should let the other score over a hundred unguarded goals in order to win the points tiebreaker. In the end, the two teams that pull off this stunt are tied in the standings -- and rather than have a second tiebreaker, the teams are declared co-champions.

I can imagine something similar happening in the WBC -- except that since the tiebreaker is based on defensive runs per inning, teams must rack up *outs*, not runs. Say instead of 10-9, 11-10, and 11-9 (which are high scoring by baseball standards), we change these to the lowest possible scores, 1-0:

Italy beats Mexico 1-0 (none out in ninth)

Venezuela beats Italy 1-0 (but in nine innings, not ten)

Also, let's assume that instead of losing to Puerto Rico, all three teams *beat* the island team. Now if Mexico beats Venezuela, there are three 2-1 teams and Puerto Rico at 0-3. In this case, the rules state that the first place team (as determined by the same run-to-inning ratio) advances to the next round outright, while the other two teams play the tiebreaker game. So the goal now is to *avoid* the tiebreaker game and advance directly to the second round.

So what must Mexico do now? If they lose they are eliminated, but if they win, they could even advance to the second round outright! We calculate as follows:

Italy: 1/18 = 0.06 runs per inning

Venezuela: *m*/18 runs per inning

Mexico: (*v* + 1)/17 runs per inning

In order for Mexico to advance outright, it must have a better ratio than Italy. So we have:

(*v* + 1)/17 < 1/18

18(*v* + 1) < 17

18*v* + 18 < 17

18*v* < -1

*v* < -1/18

Oops -- Mexico can't give up a negative score!That's This means that it's impossible for Mexico to avoid the tiebreaker game -- unless the game goes extras. It's easy to see that if Mexico pitches 10 shutout innings, both teams will have given up one run in 18 innings, and if Mexico pitches 11 shutout innings, it will have a better ratio than Italy. A 1-0 victory over Venezuela in 11 innings will give Mexico the best ratio of the three tied teams and allows them to avoid the tiebreaker.

So Mexico comes up with the following plan -- for the first nine innings, both teams simply agree not to swing at the ball. The pitchers on both teams make every pitch directly into the strike zone, and so every single batter strikes out. After both teams put up nine innings of zeros, they agree to play the game straight in the tenth. As you can see, this is the World Baseball Classic equivalent of the Quidditch match in the above link -- instead of unguarded goals, we have uncontested outs.

But why, you may ask, would Venezuela agree to this charade? Notice that Venezuela at this point is already 2-0 and is guaranteed at least a tiebreaker -- so all it has to play for is avoidance of the tiebreaker to advance outright. To do so, Venezuela needs to have a better ratio than Italy -- and here comes Mexico agreeing to put up nine zeros to improve Venezuela's ratio! And so Venezuela agrees to the farce!

There's one caveat, though. Venezuela has no incentive to strike out in the bottom of the ninth. At this point Venezuela can just swing for the fences and a 1-0 walk-off win -- this would put them in the second round outright, as their record would be 3-0. The agreement then would be to play strikeout ball for 8 1/2 innings and then play real baseball from the bottom of the ninth on. Mexico would be at a disadvantage since Venezuela has one more active at-bat -- it may be safer just to play real baseball all along and go for the tiebreaker game.

But just imagine what would happen if such a game were to be played. The fans in the stands would be bored watching every player take three straight pitches -- but stats geeks would love a game in which one team pitches a no-hitter (indeed a perfect game) for nine innings while the other team takes a perfecto into the ninth. And Italy would be furious -- they think they're guaranteed a spot in the second round with their 0.06 run ratio, and suddenly they're forced to play a tiebreaker because two teams choose not to swing at the ball. At the very least, it would force the WBC to consider changing the tiebreaker rule to prevent such a stunt from being played again.

Well, as long as the tiebreaker rule exists, we have some interesting Algebra I problems. Who would think that we could ever see (*v* + 1)/17 < 1/18 (or even 18(*v* + 1) < 17) in a real-life scenario? This question may be a bit tricky for my eighth graders though, as I never did cover inequalities at all, or even equations which require clearing fractions.

That's funny -- I said that today I want to return to Geometry, but instead I end up writing much more about an Algebra I problem. Well, let's start the Geometry right away.

Lesson 12-1 of the U of Chicago text is on Size Changes on a Coordinate Plane. If we remember from last year, "size change" is what the U of Chicago calls a "dilation." This is a good place to make my grand return to the U of Chicago, since I like emphasizing the Common Core transformations, which include dilations.

In the past, I skipped over Lesson 12-1. This is because I was mainly concerned with circularity -- dilations are used to prove some of the properties of coordinates, but right in this lesson, coordinates are used to prove the properties of dilations.

But last year, I was fed up with juggling the order of the U of Chicago text (and this year, I got in trouble trying to juggle the Illinois State text as well). This year I want to stick to the order as intended by the authors of the U of Chicago text. And furthermore, we've seen that the actual dilation problems on the PARCC and SBAC involve performing dilations on a coordinate plane -- not using dilations to prove properties of coordinates! So Lesson 12-1 is more in line with PARCC and SBAC.

Here is the main theorem of Lesson 12-1 along with its coordinate proof:

Theorem:

Let S_*k* be the transformation mapping (*x*, *y*) onto (*kx*, *ky*).

Let *P'* = S_*k*(*P*) and *Q'* = S_*k*(*Q*). Then

(1) Line *P'Q'* | | line *PQ*, and

(2) *P'Q'* = *k* * *PQ*.

Proof:

Let *P* = (*a*, *b*) and *Q* = (*c*, *d*) be the preimages.

Then *P'* = (*ka*, *kb*) and *Q'* = (*kc*, *kd*).

(1) Line *P'Q'* is parallel to line *PQ* if the slopes are the same.

slope of line *P'Q'* = (*kd* - *kb*) / (*kc* - *ka*) = *k*(*d* - *b*) / *k*(*c* - *a*) = (*d* - *b*) / (*c* - *a*)

slope of line *PQ* = (*d* - *b*) / (*c* - *a*)

Thus line *PQ* | | line *P'Q'*.

(2) The goal is to show that *P'Q'* = *k* * *PQ*.

From the Distance Formula,

*PQ* = sqrt((*c* - *a*)^2 + (*d* - *b*)^2).

Also from the Distance Formula,

*P'Q'* = sqrt((*kc* - *ka*)^2 + (*kd* - *kb*)^2)

= sqrt((*k*(*c* - *a*))^2 + (*k*(*d* - *b*))^2) (Distributive Property)

= sqrt(*k*^2(*c* - *a*)^2 + *k*^2(*d* - *b*)^2) (Power of a Product)

= sqrt(*k*^2((*c* - *a*)^2 + (*d* - *b*)^2)) (Distributive Property)

= sqrt(*k*^2)sqrt((*c* - *a*)^2 + (*d* - *b*)^2) (Square Root of a Product)

= *k*sqrt((*c* - *a*)^2 + (*d* - *b*)^2) (Since *k* > 0, sqrt(*k*^2) = *k*)

= *k* * *PQ* (Substitution) QED

At the end of this post, it's back to posting worksheets based on the U of Chicago text. This time, I post an activity from last year where students dilate cartoon characters. This activity makes more sense this year than last year since it requires using coordinates.

This lesson could actually help my eighth graders as well. We were supposed to cover dilations earlier but we ran out of time. For that matter, slope and the Distance Formula are also part of the eighth grade curriculum. I wouldn't make eighth graders perform the two preceding proofs with so many variables, but specific numerical examples are within the reach of eighth graders.

Next week is Days 122-126, so I will post Lessons 12-2 through 12-6 then. Fortunately, most of these will be cut-and-paste worksheets from last year -- and I definitely need a break next week! This is because I actually have tickets to the three games of the World Baseball Classic that are being played right here in Los Angeles, at Dodger Stadium. So cutting and pasting is all I'll have time to do!