# Ask Professor Puzzler

Do you have a question you would like to ask Professor Puzzler? Click here to ask your question! Or click the following link if you are looking for our Vortex Based Math expose.Gloria asked, "Does the popular vote not count if it's a landslide with the electoral college? How does that work?"

Gloria is asking about the presidential elections. We often talk about how much a candidate is "up" in the polls, and that's a measure of the popular vote. For example, as I write this, Secretary Clinton is, according to most polls, up by 5 to 10 percentage points. This means that 5% to 10% more voters are likely to vote for Secretary Clinton than vote for Mr. Trump.

The definition of "landslide" is pretty nebulous; one source said that if one candidate earns 60%, that's considered a landslide. Of course, with the relative unpopularity of the current candidates, that's certainly not likely. However, with Clinton leading significantly over Trump at this point, many are referring to this as a possible landslide.

But even if that's true, does it guarantee a win for Secretary Clinton? No, it does not. Because it is not the popular vote that determines the result of the election. It's the electoral college. Every state is assigned a certain number of delegates who will cast an "electoral vote" on behalf of the people in their state. For example, California, because it is such a populous state, has 55 electoral votes, but Wyoming, because it has a much lower population, only gets 3 electoral votes.

So what does the popular vote decide? It decides who each state's electors must cast their vote for. In other words, if Secretary Clinton wins California, then all 55 Californian electors cast their ballot for her. If Mr. Trump wins Wyoming, then Wyoming's 3 electors vote for him.

It's an all-or-nothing deal in all but two of the states. Maine and Nebraska have rules that allow the electors to have a split between the candidates.

So, doesn't that mean that a landslide in the popular vote would result in a landslide in the electoral college? Not necessarily. To see why, let's break it down into a simple example. Let's pretend that the United States of America consists of only two states. Semi-random selection off a map: Michigan, and its neighbor, Ohio.

Ohio has a slightly larger population than Michigan, and therefore a slightly larger number of electoral college delegates. Ohio has 18, and Michigan has 16. We'll use 2012 statistics and say that 4.9 million people in Michigan will vote in the upcoming election, and 5.6 million Ohioans will vote.

Now, to make our example clear, we're going to depart from what's likely to happen and imagine an extreme scenario for these two states deciding the election. Let's say that Michigan votes overwhelmingly for Secretary Clinton:

Michigan votes for Clinton: 4.0 million

Michigan votes for Trump: 0.7 million

Michigan 3rd party votes: 0.2 million

Let's further suppose that Mr. Trump wins Ohio, but it's a *much* closer race than Michigan:

Ohio votes for Clinton: 2.7 million

Ohio votes for Trump: 2.8 million

Ohio 3rd party votes: 0.1 million

Who wins this election? Well, Secretary Clinton gets Michigan's 16 electoral votes, because she won the popular vote there, and Mr. Trump gets Ohio's 18 electoral votes, which means Trump wins the election.

But who won the popular vote? Was it a landslide? Let's find out:

Total votes cast: 10.5 million

Total votes for Clinton: 6.7 million, or 63.8%

Total votes for Trump: 3.5 million, or 33.3%

Total 3rd party votes: 0.3 million, or 2.85%

Is this a landslide? I think any reasonable definition of landslide would agree that in this extreme scenario, Secretary Clinton's win in the popular vote was a landslide. But even so, Mr. Trump wins the electoral college vote, 18 to 16, and becomes the next president.

So what happened? What happened is that in the state that supported Clinton, she *trounced* Trump, but in the state where Trump won, it was a much closer battle. Since it's an all-or-nothing battle in both states, how *much* she won by is irrelevant to which direction the electors cast their vote.

Now add 48 more states into the mix, and imagine this sort of scenario playing itself out over and over again. Some states are close, some are a blowout, but the electoral votes are (with the Maine/Nebraska exceptions) assigned all-or-nothing regardless of how big the margin was.

Chris, from Maine, points out that this is not just a theoretical exercise: " A president has lost the popular vote but won the electoral college four times, with the most recent being only 16 years ago!"

Finally we get to the issue that a candidate is required to get a majority of the electoral votes (270). What happens if no one hits that magic number is a question for another day.

Fourth grader Alice asks, "How many diagonals are in decagon?"

Great question, Alice. But instead of giving you an answer to that question, I'm going to show you how you can figure it out for yourself. And not only that, I'll show you how you can figure out the answer for *any* polygon, even if it has 20, or 50, or 2000 sides!

Ready? Let's start with a simple example. We'll go with a polygon that has 8 sides (which is called an octagon), like this:

Now, this polygon, because it has eight sides, also has eight vertices. It seems a bit overwhelming to try to figure out *all* the diagonals, so let's just focus on one vertex. How many diagonals can you draw in an octagon, that all begin at one vertex?

The answer to that question is five. Take a look at the diagram below:

In this diagram I've chosen the diagonal to the left just below the top vertex, and I've drawn diagonals to every vertex that I *can *draw a diagonal to. Which diagonals can't I draw diagonals to? Well, I can't draw a diagonal from a vertex back to itself, and I can't draw diagonals to the two vertices next to that one (because those wouldn't be diagonals; they'd be sides!).

So there are three vertices I can't draw diagonals to from that vertex. Eight vertices total, minus the three I can't draw to, leaves five. That's important to remember: subtract 3 from the number of vertices, and you have the number of diagonals you can draw from any vertex.

But that's just one vertex! You could draw diagonals from *any* vertex, right? Sure! And there are eight vertices in our octagon, so we take the number of diagonals per vertex, and multiply it by the number of vertices: 5 x 8 = 40.

You might be tempted to think that that's our answer, but it's not. Why not? Because we've counted every diagonal *twice*! Every diagonal has two vertices, so we've counted it from one end, and we've also counted it from the other end. So to get the actual number of diagonals, we have to divide our answer by 2. 40 / 2 = 20. There are twenty diagonals in an octagon.

Let's try that reasoning on a polygon that has 2000 sides. Now, I'm not going to draw a 2000-gon for you; we'll just have to work out the reasoning without a picture.

- Pick a vertex. How many diagonals can you draw from that vertex? Answer: 2000 - 3 = 1997.
- How many vertices total? Answer: 2000
- How many diagonals counted from all the vertices? Answer: 1997 x 2000 = 3,994,000
- Oops! We've counted each diagonal twice! How do we get rid of the duplicates? Answer: 3994000 / 2 = 1,997,000

So a 2000-gon has 1,997,000 diagonals!

After doing something as crazy as a 2000-gon, a decagon shouldn't seem so difficult, right? I'll get you started, and you can finish it from here. I picked a vertex and drew diagonals from that vertex.

How many are there? Were you able to figure out how many there were without looking at the picture?

Can you figure it out from here? I hope so! Good luck, Alice.

By the way, there is a formula you can use; it looks like this: D = n(n - 3)/2. But honestly, I don't expect my students to memorize it. I expect them to remember how we reason it out here. If you understand the logic, it's even better than having a formula!

Amelia from Georgia asks, "What do time signatures in music mean?"

Great question, Amelia. I'm going to give you a little more info than you asked for; I'm going to give an answer to your question, and then I'm going to give you some examples of some unusual time signatures in musical compositions.

A time signature is a pair of numbers that you'll see, always at the beginning of a piece of sheet music, and sometimes in the middle of the sheet music as well, since some musical compositions have more than one time signature. Here's an example of a fairly common time signature in the picture to the left.

What exactly do these numbers mean? Well, the bottom number means that a quarter note gets a beat, and there are three beats in a measure. That may or may not help you understand, depending on what music terminology you understand, so let's break that down a bit.

**A Beat **is the basic unit of time in music. It's like the pulse, or the heartbeat, of a piece of music. If you're clapping along with a piece of music, then there's a good chance you're clapping on the beats. If you're tapping your feet, you're doing the same. If you hear a band playing, and they start off by saying, "1 2 3 4" before they start playing, that counting is helping to set the heartbeat of the music. If you're listening to a drummer, you'll probably hear drum strikes that aren't on the beat, but you'll definitely hear the kick and the snare on the beats. The underlying pattern in most songs is kick-snare-kick-snare over and over again (with other things mixed in). that kick-snare pattern is what sets your toes tapping.

**A Measure** can be thought of as another unit of time in music; a measure is a *set* or *group* of beats. In most music, every measure has the same number of beats, and the first beat in a group gets more emphasis than the others. Sometimes a beat in the middle of the measure will get some emphasis as well. Going back to our example of the kick-snare-kick-snare, that bass drum kick is the emphasis (you could think of it as KICK-snare-Kick-snare/KICK-snare-Kick-snare/etc.), and it repeats over and over throughout the song. In music, measures are defined by a vertical bar right through the middle of the musical staff, like this:

You can see those vertical lines before the words "world," "Lord," "come" and the second syllable of "receive." Those lines mark of groups of beats. And if you were to sing this song aloud, you would realize that the words (or syllables) immediately following those vertical lines are the ones that receive the stress or accent.

We have one more thing we need to explain in order to fully understand a time signature...note names. Take a look at the chart below:

In this chart, the notes, from left to right, are: "whole note," "half note," "quarter note," "eighth note," and "sixteenth note." Why are they called those? Because in the most common time signature (4/4 time), a whole note takes up a WHOLE measure, a half note is HALF of a measure, and a quarter note is one QUARTER of the measure. Of course, if you're not in 4/4 time, that's not true, but the name has stuck. So if you see a 4 in the bottom of a time signature that means a quarter note gets one beat. If you see a 2 in the bottom of the time signature, that means a half note gets one beat. If you see an 8 in the bottom of the time signature, that means an eighth note gets one beat.

Okay, with all of that behind us, let's go back to the original example of 3/4 time. The three means that there are three beats in a measure, and the four means that a quarter note gets a beat. So if you were clapping this one, you would do "CLAP-clap-clap-CLAP-clap-clap," and each "big" clap is the beginning of a new measure. If you were playing the drums, you might play "KICK-snare-snare-KICK-snare-snare."

### Interesting Time Signatures

The majority of music you'll listen to is written in 4/4 time (four beats per measure, a quarter note gets a beat). Some will be 3/4 (which is often referred to as a waltz), or 6/8 time (six beats per measure, an eighth note gets a beat - this is the common meter for Irish jigs). Once in awhile you'll hear something that's a bit different.and I've selected a few that you can listen to on youtube.

**Of the Father's Love Begotten **- since I've already brought up a Christmas song, here's another - an old old Christmas hymn. If you look at sheet music for this one, you'll likely find that the publisher hasn't even bothered trying to put a time signature on it, and has more-or-less given up on doing measure markings; the measure markings are actually just phrase markings, and every phrase is a different length. The link I've provided has no singing, but does show the sheet music.

**Waltz from Tchaikovsky Symphony #6** - This one is called a waltz, but it is not in the normal 3/4 time; it's in 5/4 time instead. This means that it has 5 beats in a measure, which gives it a slightly off-kilter feeling because our brains aren't used to processing that kind of rhythm. It's a beautiful, elegant piece of music.

**Finale from Stravinsky's Firebird Suite** - My introduction to this piece of music was in high school when I participated in the All-State music festival. In the linked video, the section in an unusual time signature starts at 2:00, but it's interesting to listen to the lead-in to that, so it's worth listening to the whole video. The unusual time signature is 7/4 time. I remember looking at the sheet music and thinking, "What in the world?" because every measure was subdivided into groups of 3 beats and 4 beats - presumably to help us get the count right!

**The Dargason from Holst's Saint Paul Suite** - This one I played in university orchestra, and oh, was it fun! What makes this piece so unusual is that there are two melodies going on simultaneously, and they are in different time signatures. One melody is "Greensleeves," which is in 3/4 time, and the other is a jig, which is in 6/8 time. Now, as a mathematician, I notice that these two fractions - 3/4 and 6/8 are equivalent. This means that the measure marks fall in the same place for each melody, but what happens in the middle of the measure is quite different.

**Here Comes the Sun** - In general, the Beatles were notorious for shifting into unusual beat patterns. I've picked "Here Comes the Sun" as an example. You can feel the 4/4 beat through most of the song, but when they get to the bridge, everything seems to go haywire. I haven't looked at the sheet music for this one, but Wikipedia says that the bridge goes through the following rhythm structure: 11/8 + 4/4 + 7/8.

Every year that I teach Physics I run into a small problem with the labs my students do. Here's the problem. My students will collect data (and the issue happens most often when they're timing something) and then they'll look at their data and say, "Wow! One of those points is way off from the others!"

For example, their data points might be the following times in seconds: *4.2, 4.6, 4.3, 4.8, 7.2, 3.9*

As they look at those data points, they'll recognize that most of them are in the vicinity of 3.9 to 4.8, but there's one value (7.2) that is way off from the others. Obviously something went drastically wrong with that trial!

So what do I tell them? I'd like to tell them, "Use your statistics knowledge to determine the outliers." Outliers are points like 7.2, which are clearly outside the reasonable range. That's what I'd like to tell them, but I don't, because I know perfectly well that many high school science students have not yet learned enough statistics to calculate outliers.

And I have no intention of spending my Physics class time teaching statistics!

So what do I do? I say to them, "If it looks ridiculous, dump it."

Now, that's a quick approach, and it saves a lot of time, but it's not really a "good" approach (although I'm not the only science teacher that does it!). Why is not a good approach? Because good scientists try to avoid using intuition and guesswork when selecting their data. That kind of fuzzy thinking can result in people massaging their data to make it say exactly what they want it to say.

So this year, I decided it was time to resolve that problem. Not by teaching my students how to find quartiles, interquartile ranges, and outliers. No...I decided to create a calculator that they could enter their data in, and have the calculator list their outliers.

And, while I was at it, since I realize these kinds of tools can be very helpful for both teachers and students, a whole section of statistics calculators was added to the site. These include calculators for various means (arithmetic, geometric, and harmonic), as well, as standard deviation, variance, and other useful statistical quantities.

You can find the calculators here: **Professor Puzzler's Statistics Calculators**.

Twelfth grader Abbey wants some help with the following: "Factor x^{6} +2x^{5} - 4x^{4} - 8x^{3} + x^{2} - 4."

Well, Abbey, if you've read our unit on factoring higher degree polynomials, and especially our sections on grouping terms and aggressive grouping, you probably realize that a good way to attack this problem is to try grouping the terms. Hopefully, you tried something along those lines. Here's one method (I'll get you started on another method at the end of this post):

(x^{6} + 2x^{5}) - (4x^{4} + 8x^{3}) + (x^{2} - 4).

That looks promising, since every group can be factored:

x^{5}(x + 2) - 4x^{3}(x + 2) + (x - 2)(x + 2).

Great! every group has an (x + 2) term in it, so we can factor that out:

(x + 2)(x^{5} - 4x^{3} + x - 2)

I'm going to attempt grouping on that second polynomial:

(x + 2)[(x^{5} - 4x^{3}) + (x - 2)]

(x + 2)[x^{3}(x^{2} - 4) + (x - 2)]

(x + 2)[x^{3}(x - 2)(x + 2) + (x - 2)]

And once again we have a common factor in our groups - this time it's (x - 2).

(x + 2)(x - 2)[x^{3}(x + 2) + 1]

(x + 2)(x - 2)(x^{4} + 2x^{3} + 1)

Now we have something that can't easily be grouped, because it has 3 terms. So I decided to try something crazy: I added in x^{2} and subtracted it back out again. Why? Take a look:

(x + 2)(x - 2)(x^{4} + 2x^{3} + x^{2} + 1 - x^{2})

(x + 2)(x - 2)[(x^{4} + 2x^{3} + x^{2}) - (x^{2}^{}- 1)]

(x + 2)(x - 2)[x^{2}(x^{2} +2x + 1) - (x + 1)(x - 1)]

(x + 2)(x - 2)[x^{2}(x + 1)^{2} - (x + 1)(x - 1)]

My groups each contain an (x + 1), so I can factor that out:

(x + 2)(x - 2)(x + 1)[x^{2}(x + 1) - (x - 1)]

(x + 2)(x - 2)(x + 1)(x^{3} +x^{2}^{ }- x + 1)

At this point, I tried a couple different groupings, and nothing seemed to work to factor (x^{3} +x^{2 }- x + 1). That doesn't mean it's not factorable, but I got a little suspicious, so I pulled out the Rational Root Theorem, which tells me that if that thing has any rational zeroes, they have to be either 1 or -1. Since neither of those are zeroes of the polynomial, that means it's unfactorable over the rationals. Therefore, the final answer is:

**(x + 2)(x - 2)(x + 1)(x ^{3} +x^{2}^{ }- x + 1)**

Just a couple notes on this, before I'm done:

- I enjoy the process of trying to do grouping (and what I call "aggressive grouping") on polynomials in order to factor them. I think it's a good mental exercise. Nevertheless, I should point out that according to the Rational Root Theorem, if this polynomial has any rational zeroes, they are any of the following: -1, 1, -2, 2, -4, 4. Trying out those to see which work, and then using Synthetic Division to divide out the factors will guarantee you a solution, even if you can't figure out groupings to do.
- There's another way to start off the problem (which is actually what I did the first time I solved the problem), which involves rearranging the terms before grouping: (x
^{6}- 4x^{4}) + (2x^{5}- 8x^{3}) + (x^{2}- 4). Grouping this way has the advantage of pulling out (x + 2)(x - 2) all in one step, and going straight to a 4th degree polynomial. It's nice to see that the problem can be done in multiple ways, don't you think?