The question came from a teacher in 2003:

Defending the Mean I've been asked by one of my students to explainwhy it makessense to add the sum of the numbers in a data set and then divide by the quantity of numbers when finding the average or mean.

Doctor Achilles answered by going beyond the question about the mean, and comparing its purpose to the median and the mode:

Hi Leigh, Thanks for writing to Dr. Math. That's an excellent (and thoughtful) question. I believe thatwhen you take time to work through the math that is going on, it is intuitive, but I'm not sure if I can give a more principled argument. Let me work through what I mean. Let me explain what I mean by this using students and apples. Let's say that we have a set of students, and each student has some number of apples. Some students may have more apples than others. But I want to knowhow many apples a typical student might have. One way to think about this "typical" student is this: If another student walks into the room, how many apples would you guess she has? Let's take a very simple case: we have 20 students and every single one of them has 8 apples. If I say another student just arrived,how many apples would you guess she hasjust based on what you know about the 20 students who are already there? You'd probably guess she has 8, wouldn't you? You might not be shocked if I told you she has 7 or 9, but if it turns out she has 17 or 20, you would probably say that there must be something different about her that caused her to bring more apples.

Each kind of “average” reflects a different perspective on this idea of “typical”.

There arethree intuitive waysI can think of to define this "typical" student. The first is to look and see if a lot of the students have the same number of apples. For instance, if we have 4 students and 3 of them all have 7 apples and the 4th student has 9 then it might make sense to say that a typical student has 7 apples. This, as you probably know, is called the "mode". For a little more information about modes which is not really relevant to this discussion but may be interesting to you or your students, check out this page: More Than One Mode? http://mathforum.org/library/drmath/view/61375.html

We’ll be looking at that page in a later post. But here are our students:

The mode is literally “typical” (that is, what *most* students are like) – if there is a mode at all. It isn’t good at taking others into account.

Another way you might define the typical student is to line up all the students in order from the one with the fewest apples to the one with the most, and then say thatthe one in the middle of the lineis the "most typical". I personally find this the least intuitive of the 3 ways to define a typical student, but it is certainly a way you could do it. As you probably know, this is called the "median". This is really, by definition, not so much an assessment of what is typical of the population as a whole as just what is actually true of the one individual in the middle of the population.

So what about the mean?The advantage that the mean has over the medianis that it tells you something about the population beyond just what is at the middle.The advantage that the mean has over the modeis that the mode only works if there is one number which occurs more than any other. Also, modes only tell you about the largest part of the population but ignore any minority groups. Let's start with an easy case forwhy we use the mean. Let's say that there are 2 students, one with 7 apples and the other with 9. What is a typical student in that group? Well, it seems intuitive to me (before I really even do any math), thata typical student should be halfway betweenthe two students in our population. For me, it just plain makes sense to say that a typical student has 8 apples in this set. No other number represents the set well. You could say 7 is typical, but that ignores the student with 9 apples. You could say 9 is typical, but then you have an identical problem. You could say 10 or even 0 or 20, but none of these numbers are even close to 9 or 7 and you would be downright nuts if you said any of them were typical of this set. You could maybe try saying 7.5, that doesn't completely ignore the student with 9 apples, but it means you're paying more attention to the student with 7. Similarly 8.5 doesn't completely ignore the student with 7, but it pays more attention to the student with 9.

Like the median, the mean is literally “in the middle” in this example. (In fact, we would use the same number for the median.)

Someone with 8 apples has just as many more than the 7 as she has less than the 9.

This is easily seen on the number line:

Here, the mean is right where a seesaw with those two dots would balance.

Let's move to a slightly more difficult case. Let's say we have two students with 7 apples each and one student with 10 apples. What is typical in this group? Well, you could say 7 is typical (the mode) but that ignores completely the student with 10 apples. You could say 8.5 is typical (halfway between 7 and 10) but there are two students with 7 apples and only one student with 10 apples, so shouldn't wepay more attention to the 7 apples? What about 8, is that a reasonable compromise? Sure, it is only 1 away from 7 and it is 2 away from 10; it make perfect sense for it to betwice as far from 10 as it is from 7because there are twice as many students with 7 apples as there are with 10 apples.

Here are our students, with our proposed mean:

Again, 8 fits in well.

If you know any physics, you can see that this, again, will balance at the mean.

You can make up more examples, but if you work through what seems like a fair way to find a typical student in each group, I bet you'll always come up with the mean.

This has all been pretty subjective. But now we come to what I consider the best explanation:

Another way to think of the mean is this: if every student put his or her apples into one big pile (take the sum of all the apples) and then we passed out anequal number of apples to each student(divide by the number of students), how many apples would each student have? Exactly the mean! So the mean is a good way to think about what the group would look like if everyone redistributed their apples so that all the students had the same number.

And this is exactly what we can up with in our example: $$\frac{7+7+10}{3} = \frac{24}{3} = 8$$

Equivalently, imagine the one with 10 giving an apple to each of the others. Then they are equally distributed, and each one has \(\frac{1}{3}\) of the total. There are 24 apples in all (7 + 7 + 10), and each has \(\frac{1}{3}\) of them:

The rest of this post will largely be different ways to see this same argument.

By the way, were you wondering about the mean of our original set of numbers, 7, 7, 7, and 9? Here it is:

Using the formula, the mean is $$\frac{7+7+7+9}{3} = \frac{30}{4} = 7.5$$

Leigh replied,

Thank you very much for your quick response. That will be helpful tomorrow in my math class and to answer my student's questions. In college,we are taught how to teach to calculate the mean, but not why, so thank you very much.

“Why”, you may have observed, is what Math Doctors live for!

The next question is from 2001:

What Does Average Mean? Can someone explainwhat average is? For example what is the average rainfall in New York? I understand the formula needed to find the average, butwhat is the meaningof average?

I responded, focusing on the idea of rainfall:

Hi, John. Sometimes "average" just means "normal." Other times it refers to some number "in the middle"; there are several different ways to define an average. But the usual average, which you are familiar with, called the "arithmetic mean," is calculated by summing all the values of something and then dividing by the number of values you used. What this really means is that you arefinding a single number that you could usein place of all the different numbers and still get the same total. For example, suppose that the rainfall one day is 2 inches, the next day 0 inches, and the next day 1 inch. The total rainfall over the three days is 3 inches, so it is just as if there had been 1 inch each day. So the average daily rainfall over that period is 1 inch.

Do you see thatan average "smooths out" the numbers? It's as if there were two inches of dirt on one block of sidewalk, none on the next, and one on the third. If I spread all the dirt out evenly, so it was all the same level, there would be one inch over the whole sidewalk.

(I switched to dirt because water doesn’t pile so well.)

In the same way,if we spread the rain out evenlyover all three days, there would be one inch every day. Of course, everything isn't "just as if it rained an inch every day"; then you could never have a picnic in Central Park! But justfrom the perspective of the total amount of rain, it's the same. And that's why there are different kinds of average: if something else is important other than the total, then you can get a different "average."

We’ll be seeing more kinds of mean next time, which apply when something other than the sum is meaningful.

The picnic idea is a reminder that any kind of average reduces lots of data to a single number, and loses any information about how the data are distributed. *How* it piles (the distribution) can be just as important (or more so) as how much there is overall.

It's also important to ask what data are represented by the average you are talking about. Is the "average rainfall in New York" theaverage per dayover some particular number of days? Is it the averageover a whole year? Is it theaverage per yearover a number of years? Or the average amounton December 5 over several years? The actual amount will vary with the season as well as from year to year, and you can average the data to eliminate either variation, or both, depending on what your goal is.

Average? Of what? Over what time period? These are important questions to ask.

Continuing the same idea, here is a question from a teacher in 2003 asking for a way to use manipulatives:

Teaching the Concept of Average How can I use objects to teach averages to students?

I answered, first referring to that last answer, and then turning it into a demonstration:

Hi, Ridya. For a discussion of what averages are, see our archives: What Does Average Mean? http://mathforum.org/library/drmath/view/52809.html As I explain there, the essential idea of averaging is "smoothing numbers out." You might introduce the idea by giving each student a different number of, say, blocks or balls or cards, and asking how many each of them has. Write those numbers down, and point out how they vary. Then have themwork together to find a way to share them equally. (You may or may not have chosen the total number of objects such that they CAN be shared exactly.) Eventually they should all have about the same number of objects, and you can discuss with themhow they could have decided ahead of timehow many each of them should have. The answer, of course, is to find the total number of objects and divide by the number of students. The resulting number - the number that each has if they share equally - is called the average.

There are several strategies the students might invent. One, as I’ve suggested, is to pile the blocks together and then hand them out one by one; another is to have the student with the most give some to the one with least, and repeat until they are nearly equal. The first of these is a better way to understand the calculation, and is also often more efficient. But sometimes just moving a few blocks around may be quicker! That is shown next:

You can also do this as a smaller-scale demonstration. Make piles of blocks of different heights, like H H H H H H H H H H H H H H H H H H H H ------------------------- Then move blocks from the higher piles to the lower ones to make all the piles equal: H H H H H H H H H H H H H H H H H H H H ------------------------- You have just found the average: on the average, each pile had 4 blocks in it. And how can you find that number if you just have the numbers 5, 6, 3, 2, and 4 from the original piles? You add and then divide. So, yes, division is central to finding an average. But addition is the other half of the work. Does this help clarify the relationship?

Even though we didn’t make a single pile, the result again suggests the calculation: The end result is 5 piles of 4, so the total count is \(5\times 4 = 20\); and we can see that if we knew the total, we could find the average by dividing \(20\div 5 = 4\).

(Incidentally, if you are wondering why I commented on the role of division, that’s because in editing for the archive, the question was shortened. In the original, Ridya expressed confusion about the difference between average and division.)

Ridya replied,

Good Day, Dr.Math! Thanks for your explanation as it really helps me out to solve my problems regarding average. I will use your ideas to introduce this topic to my students. Thanks again. Ridya

Our last question is from 2008:

Average Speed and How It Relates to Average in General Theaverage speedof an object is distance/time. We are taught that the average is found by adding up the number of items and dividing by how many there are. How does this average relate to the average speed or are they unrelated?

On the surface, these seem like very different things: a ratio of distance to time, vs. adding numbers and dividing. Why are they both called “average”? It turns out that they are the “continuous” and “discrete” versions of the same concept.

I answered this:

Hi, Brian. Yes, these ideas are very closely related. The essential idea of "average" in both senses is "a uniform value that could be put in place of all values in a collection, without changing the net effect". In the case ofaveraging numbers(specifically what is called the arithmetic mean), this means that you want to findwhat number they could all be replaced by, and still have the same sum. As an example, if 10 people all had different amounts of money and wanted to share it equally, they would put it all together (adding the amounts) and then each take an equal share (dividing by the number of people). Then they all have the same amount, but the total is the same as it was originally.

This is the same idea we’ve been looking at: $$\frac{\text{Total value}}{\text{Total count}}$$

In the case ofaverage speed, suppose you have been driving at varying speeds during different parts of a trip, but you want to know atwhat speed you could have driven to take the same amount of time, if you had driven at a constant speed. To find out, you take the total distance driven, and divide by the total time it took. If you had driven at that speed throughout the trip, you would have arrived at the same time. In a sense, we have divided the distance equally among all minutes of the trip.

This time, it’s $$\frac{\text{Total distance}}{\text{Total time}}$$

Note thatin both cases you are dividing a total by a total--the total amount of money by the total number of people, and the total distance by the total time. In fact, if you drive distinct segments of your trip at different constant speeds, it turns out that the average speed is theweighted averageof the individual speeds: you multiply each speed by the time for which you drove that speed (which is called a weight) and then divide by the sum of the weights (the total time).

Weighted average is another concept we’ll be looking at later in this series. But next week, we’ll look at different means: arithmetic mean, geometric mean, harmonic mean, …

]]>Here is the problem (I’ve replaced the original image of the problem with my own, which is cleaner):

Hi Doc!

I am having a difficult time trying to solve one problem with my homework, and I’d really appreciate some help with it.

The question goes as follows:

“Point O is the center of the circle, AB is perpendicular to BC, AP = AD, and AB has length twice the radius of the circle. Prove that AP^2 = PB*AB”

I’ve attached the graph as well:

This problem is from a geometry class I’m taking, and the teacher said this one was particularly difficult, and it definitely seems to be. We’ve learned the

interior chord theorem, if that’s what it’s called, where if there were 2 chords, AB and CD that met up at P, AP*BP = CP*DP. Also, we’ve learned anexterior “chord” property? I’m not sure what to call it, but it’s when 2 chords AB and CD don’t intersect inside the circle but outside the circle by extending the chords outwards to meet at point P. If that is the case, then PA*PB = PC*PD. Also, if there is a line segmenttangentto the circle PF, then PF^2 = PA*PB, or PF^2 = PC*PD.We’ve learned a lot more than this, of course, but this is all the new things we’ve learned this week that is for the homework, and maybe a little bit of what we’ve learned before.

This is a very well-presented question! It illustrates well most of the recommendations I made in How to Ask a Good Question. When we pick up the question again below, we’ll see him show his work.

Before continuing with his question, let’s look at the theorems he has listed (which is an excellent thing to do, to give us context!). A good compact source I like to give students who don’t know these theorems is MathBitsNotebooks’s Rules for Chords, Secants, and Tangents in Circles. Eric’s “Interior Chord Theorem” is the **Intersecting Chords Theorem**:

Eric’s “Exterior Chord Property” is the **Secant-Secant Theorem**:

In both, the equation is \(ab = cd\), where *b* and *d* in the second theorem cover the entire segment shown.

Eric’s tangent property is the **Tangent-Secant Theorem**:

This is really the same fact, but *a* and *b* have joined to become a single segment *t*.

Together, these three theorems constitute Steiner’s Power-of-a-Point Theorem.

Now Eric tells us what he has tried (again, an excellent example of how to ask us for help!):

I’ve been trying to figure out just what the problem wants me to find out since the question is asking something very obscure and unconventional, almost like it’s just coincidence that AP^2 = PB*AB. So,

I decided variables are the way to go for this problem, and just use algebra to prove it.I thought of maybe using

xas a placeholder for AD and AP, but didn’t really do anything with it, or maybe usingrfor radius as a variable too. I do know that AB is 2r, so I tried substituting that into equation AB = AD*AE to get AD = -(AD^2/2r), and I’m not sure what to do with that strange value. I also attempted to make a triangle out of the graph, by connecting A and C, yielding AC = 2rsqrt(2), but I’m not sure how that would help. Aside from these few thoughts, I’ve also just drawn lines and relationships all over the graph, to try and see if one line or angle equality would solve it all, but to no avail.Could you try and guide me in the right direction, please? I’m very confused …

Best Wishes,

Eric

The use of algebra is indeed appropriate, though the algebra is clearly wrong. The introduction of a right isosceles triangle in an interesting observation, though it probably won’t help:

I replied:

Hi, Eric.

I think you made a good start, and just tripped over a little mistake.

I myself took r as a constant, and let x = |AD|, as you suggested, then

tried to solve for x, using triangle ABO.On the other hand,

what you started would lead to something equivalent, if you did it right. You said,I tried substituting that into equation AB = AD*AE to get AD = -(AD^2/2r).

I presume you meant AB

^2= AD*AE; in any case, I don’t see how you got that result, which clearly makes no sense. I get a quadratic equation in x. Perhaps you should show me your work there.But once you have the right equation,

you may not need to actually solve for x. The first thing I had done was to writewhat we want to prove in terms of r and x. (General principle: write down what you are given, and what you want to prove, before you do anything else, so you’ll recognize the latter when you run across it.) In fact, I simplified that desired result a bit, in effectworking backward from the goal. That helped a lot.See what you discover!

(By the way, when a chord is extended to a line, that line is called a

secant.)

Here I hinted at a place to start, using triangle ABO; I also encouraged Eric to continue with his method, and below I’ll be showing how both approaches work. I also gave some important general advice for solving complicated problems, including proofs. We’ll see it in action as we proceed.

I realize now, on looking at the image Eric had sent showing the problem (which I omitted above because it had been written on), that he had shown his work there: $$AB = AD\cdot AE\\2r=AD(AD+2r)\\2r=AD^2+2rAD$$ From there, he made some silly mistakes by misreading the equation; but since this started with an error, it wouldn’t have been right anyway. The correct equation would have been $$AB^2 = AD\cdot AE\\(2r)^2=AD(AD+2r)\\4r^2=AD^2+2rAD\\AD^2+2rAD-4r^2 = 0$$ which could be solved by the quadratic formula. (He also didn’t identify AD as *x*, as we are now doing.)

Eric responded:

Hey Doctor Peterson,

I tried doing what you said, by using the triangle to solve for x, and wrote what we needed to prove, but I’m not sure what to do next? I mean I think I tried to come across the answer… Did I just simply make a calculation mistake?

On the left, he has written the goal, \(AP^2=PB\cdot AB\), and expressed it in terms of the variable *x*, then expanded it, with the only mistake being in the last line. In the middle, he did that again, but using the given fact that \(AB = 2r\). On the right, he played with the right triangle ABO that I had mentioned, which leads to something useful, then went astray.

I answered:

Here are a few comments:

- In the lower left, you have (x – PB)^2 = 0. This is wrong; what do you get when you expand that?
- I would express PB as 2r – x; you want to express everything in terms of r and x.
- On the right you get x^2 + 2rx – 4r^2 = 0. This is correct. Stop there; you don’t want to go back to using PB.
- You got that using my right triangle ABO; you could also get it from the secant-tangent theorem you used before. (In your crossed-off work at top center, you did this but didn’t use the fact that x + PB = 2r.)
You’re getting close.

Here is the figure again, emphasizing triangle ABO and the variables:

The Pythagorean Theorem applied to ABO yields $$r^2 + (2r)^2 = (x + r)^2\\r^2 + 4r^2 = x^2 + 2rx + r^2\\x^2 + 2rx – 4r^2 = 0$$ which he had on the right; equivalently, at the top he had used the tangent-secant theorem to write what could have become $$AB^2 = AD\cdot AE\\(2r)^2 = x(x + 2r)\\4r^2 = x^2 + 2rx\\x^2+2rx-4r^2 = 0$$

Meanwhile, on the left, he had found (without using the value for PB) that the goal is $$AP^2 = PB\cdot AB\\x^2 = PB(x +PB)\\x^2-PBx-PB^2 = 0$$ which, after substitution with \(2r-x\) yields $$x^2 – (2r-x)x – (2r-x)^2 = 0\\x^2-2rx+x^2-4r^2+4rx-x^2 = 0\\x^2+2rx-4r^2=0$$ We could have made the substitution from the start: $$AP^2 = PB\cdot AB\\x^2 = (2r-x)(2r)\\x^2 = 4r^2-2rx\\x^2+2rx-4r^2 = 0$$ Either way, we have shown that the given facts are equivalent to the goal. This is not yet in the form of a proper proof (working from the givens to the goal), but we have a complete connection.

He replied:

Oh! I didn’t realize (for some reason…) that PB could be written as 2r – x. After rewriting the equation I ended up with x^2 + 2rx – 4r^2, which lined up with one of the equations for the triangle side lengths! Thanks for helping me through proving this question!

I answered:

Yes, it was important in my own work to label AB as 2r very early, so the fact that AP + PB = 2r was visible. And it was also important to have studied the goal or target thoroughly so I’d notice it. The other main feature was to write both the givens and the goal in terms of the same variables, so they would look alike. Making details visible is key.

This is a good example of the technique of working from both ends toward the middle, which is discussed here:

Let’s write it up now as a proof, by rearranging what we did above, reversing the path to the conclusion:

Given: Point O is the center of the circle, AB is perpendicular to BC, AP = AD, and AB has length twice the radius of the circle.

To prove: \(AP^2=PB\cdot AB\)

Let \(AP = AD = x\), and \(OB = OD = OE = r\).

We are told that \(AB = 2r\); and since \(AB = AP + PB\), \(PB = AB – AP = 2r – x\).

By the tangent-secant theorem, $$AB^2 = AD\cdot AE\\(2r)^2 = x(x + 2r)\\4r^2 = x^2 + 2rx\\x^2 = 4r^2 – 2rx\\x^2 = 2r(2r – x)\\AP^2 = AB\cdot PB$$

Q.E.D.

(Incidentally, this way of first working from both ends to the middle, and then turning your work “inside-out”, is also demonstrated, for trigonometric identities, in Different Ways to Prove a Trigonometric Identity.

There are probably several very different ways to prove this; let me know if you find one!

]]>Our first question is from 1997:

Mean, Median, Mode, Range What are definitions for mean, median, mode, and range?

Several of the questions here will mention range; I’m going to omit that from most answers, to narrow our focus.

Doctor Scott answered with straightforward definitions by means of an example:

Hi Jamie! Let's define each of the words and give an example. Consider the set of numbers 80, 90, 90, 100, 85, 90. They could be math grades, for example. TheMEAN is the arithmetic average, the average you are probably used to finding for a set of numbers - add up the numbers and divide by how many there are: (80 + 90 + 90 + 100 + 85 + 90) / 6 = 89 1/6. TheMEDIAN is the number in the middle. In order to find the median, you have to put the values in order from lowest to highest, then find the number that is exactly in the middle: 80 85 90 90 90 100 ^ Since there is an even number of values, the MEDIAN is between these two, or it is 90. Notice that there is exactly the same number of values ABOVE the median as BELOW it! TheMODE is the value that occurs most often. In this case, since there are 3 90's, the mode is 90. A set of data can have more than one mode.

Observe that the **mean** is a “center of gravity” takes into account each individual value, so it is pulled strongly toward the extreme:

I represent it with a triangle representing a fulcrum on which the data are balanced.

The **median** ignores the actual values, and just takes into account their order, so that it is “in the middle” in a different sense:

The **mode** pays attention only to the single most common value (if there is one):

When we use the word “average” by itself, we typically mean the mean. Next week, we’ll look into some details about it; we’ll also see that just as there are different “averages”, there are different “means”!

We have discussed elsewhere some subtleties in the definition of the median. We will eventually have a similar in-depth look at the mode.

We have previously looked at differences between these statistics in Mean, Median, Mode: Which is Best?, which focused on choosing the most appropriate measure for a given situation.

For more about what each of the “averages” means, and how they differ, we can turn to this question from 1998:

Mean, Median, and Mode I understand themedian, but I am having trouble understanding themode. Can you please explain?

Doctor Anthony answered, explaining all three:

If you have a set of numbers, say theages of pupils in a group, then there are 3 different ways of finding a single number to represent the whole group. The most common is the 'mean' or average. For themeanyou add up all the ages and divide this total by the number of pupils. The second way you could find a single representative number is to arrange all the pupils in a line in ascending order of age, with the youngest on the left and oldest on the right. You then go to the person standing in the exactmiddleof this line and find his/her age. This will be themedianage. If there is an even number of pupils you will not have a single person at the midpoint, so you will take themiddle pairand give the average of their ages as the median age. The third way to find a single representative number is to group the pupils by age, so you could have 5 pupils of age 10, 8 pupils of age 11, 14 pupils of age 12, 7 pupils of age 13, 2 pupils of age 14 and 3 pupils of age 15. Looking at this distribution of ages you see that thebiggest groupis those of age 12, so you say the 'mode' of the distribution is 12. In short, themodeis the most frequently occurring value. If no value occurs more than once then you don't have a mode. Sometimes two values will occur at an equal but greater frequency than other values, and in this case we say that the distribution is bi-modal.

As we’ll see, the main weakness of the mode is that there may not be a mode at all, or there may be more than one.

Here’s another question from 1998 that asked about one example, and got a little more:

Finding Mean, Median, Mode I have been out of school for 8 years and I am trying to get my diploma. On one of my practice tests it said to get themean, median, and modeof scores. However in my textbook there are no examples to show me how to do it. The scores are: 100, 78, 93, 84, 91, 100, 82, 79. I would greatly appreciate if someone would show me how to do them.

Here are Adam’s data, placed on a number line:

Doctor Sam answered, including helpful comments:

Adam, Themean of the scoresis another name for their average. Just add them up and divide by the number of scores: mean = (100+78+93+84+91+100+82+79)/8 = 88.375 Themedian of the scoresis "the number in the middle" when the scores are sorted in order. In your example: 100 100 93 91 84 82 79 78 If there is aneven number of scores(as in your example) there is no number in the middle so the two numbers in the middle are averaged: median = (91 + 84)/2 = 87.5 The mean of the numbers can be misleading. If I tell you the mean income in my neighborhood is 1 million dollars a year you might think that I am wealthy. But maybe the neighborhood is a poor one with one very rich person making many millions of dollars a year.The average will be large because one number is very large.The median and the mean together give a better idea of the spread of the numbers. If there were one really wealthy person then the median income would be quite low. In your case the median is quite near to the mean, which suggests that the spread is probably evenly balanced.

The comments here are that the mean is highly affected by “outliers”, values that are far from the others. This is revealed when the median is very different (as in the first example in this post); in the present example, the numbers are distributed fairly evenly.

Here are the three averages on the number line:

Themodeof a set of numbers is the number that occurs most often. In your example: mode = 100 The modemay help to correct false impressionsif you know the mean and the median but don't actually see the data. For example, if I tell you the mean of four numbers is 50.5 and the median is 50 you may think that the four numbers are close to one another, like the numbers in your example. In fact, however, I am thinking of: 101 99 1 1 The mean is (101 + 99 + 1 + 1)/4 = 202/4 = 50.5 and the median = (99+1)/2 = 50. But the mode here is 1. If you know that, you will know that 1 is repeated more than once in the data and, since the mean is near 50 with only four numbers, that might give you the idea that the data include several large numbers and several small numbers rather than four numbers close to one another.

As we’ve seen, the mode doesn’t always even exist, and therefore is not very useful; yet there are times when it can be very revealing. In Adam’s problem, however, the mode really tells us nothing! It just happens that two values are the same, and if one of the 100’s were changed to 99, there would be no mode at all.

The next question, also from 1998, is not about “what”, but about “why”:

Range, Mean, Median, and Mode I have some questions that you may want to answer for me: 1. Why do we have to study range, mean, median, and mode? 2. Could you help me understand them more? 3. How is it going to help me later in life?

Doctor Stacey took this one:

Hi Stephanie! Thanks for writing Dr. Math. Mean, median, and mode are all types of averages, although the mean is the most common type of average and usually refers to the _arithmetic mean_ (There are other kinds of means that are more difficult). Thearithmetic meanis a simple type of average. Suppose you want to know what your numerical average is in your math class. Let's say your grades so far are 80, 90, 92, and 78 on the four quizzes you have had. To find your quiz average, add up the four grades: 80 + 90 + 92 + 78 = 340 Then divide that answer by the number of grades that you started with, four: 340 / 4 = 85. So, your quiz average is 85! Whenever you want to find a mean, just add up all the numbers and divide by however many numbers you started with.

Here is the mean; you may notice something special about the data:

The term “arithmetic mean” is used to distinguish “the” mean from other means we’ll be looking at next week.

But sometimes the arithmetic meandoesn't give you all the information you want, and here is where your first and third questions come in. Suppose you are an adult looking for a job. You interview with a company that has ten employees, and the interviewer tells you that theaverage salaryis $200 per day. Wow, that's a lot of money! But that's not what you would be making. For this particular company, you would make half of that. Each employee makes $100 per day, except for the owner, who makes $1100 per day. What? How do they get $200 for average then?! Well, let's take a look: Nine employees make $100, so adding those up is 9 x 100 = 900. Then the owner makes $1100, so the total is $1100 + $900 = $2000. Divide by the total number of employees, ten, and we have $2000/10 = $200. Because the owner makes so much more than everyone else,her salary "pulls" the average up.

The owner’s salary is an “outlier”, which was mentioned before:

Only the mean is affected by the outlier! The single outlier is 9 times as far from the mean as the others. Although the outlier doesn’t pull the mean too far, it does make the mean deviate from what is typical.

A better question to ask is, "What is the _median_ salary?" The median is the number in themiddle, when the numbers are listed in order. For example, suppose you wanted to find the median of the numbers 6, 4, 67, 23, 6, 98, 8, 16, 37. First, list them in order: 4, 6, 6, 8, 16, 23, 37, 67, 98. Now, which one is in the middle? Well, there are nine numbers, so the middle one is the fifth, which is 16, so 16 is the median.

We don’t need a number line to show the median:

By the way, the way I commonly find the middle is to divide the count by 2 and round down. Then I know there will be that many on each side. In this case, \(9\div 2 = 4.5\), which rounds down to 4, so there are 4 below and 4 above the median.

Now, what about when there is aneven number of numbers? Look at the quiz grade example again: 90, 80, 92, 78. First list the numbers in order: 78, 80, 90, 92. Thetwo middle onesare 80 and 90. So do we have two medians? No, we find the mean of those two: 80 + 90 = 170, and 170 / 2 = 85. So 85 is the median (and in this case the same as the mean)!

Here we see that the mean and median are the same because the data are symmetrical:

Here when I divide 4 by 2 I get 2 exactly, so I don’t need to round down; and again, there are 2 numbers on each side of the median, which this time is between the two sets of two.

What about outliers?

Now look at thosesalariesagain. To find the median salary, we look at the salaries in order: 100, 100, 100, 100, 100, 100, 100, 100, 100, 1100. This is an even number of x salaries, so we look at themiddle two. They are both 100, so the median is $100. That's much better at telling you how much you'll make if you accept the job.

We saw this above:

Again, I divide 10 by 2 to get exactly 5, so there are 5 on each side of the middle; the fifth and sixth numbers have to be averaged.

But the median doesn't always give you the best information either. Suppose you interview with a company that has 10 general employees, 7 assistants, 3 managers, and 1 owner. For this company, themeansalary is $400, and themedianis also $400. But you are applying for the position of general employee, whose starting salary is $100! Why are the mean and median so far away? Well, the 10 general employees each make $100. The 7 assistants each make $400, the 3 managers each make $900, and the owner makes $1900. If you do the math to find the median or mean, $400 is the answer (try it!). So what can you do?

Here are these employees’ incomes on a number line:

And here is the calculation for the median:

and for the mean: $$\frac{10\times 100+7\times 400+3\times 900+1900}{21} = \frac{8400}{21} = 400$$

(Here rather than list all the numbers as I did for the median, I used multiplication as a shortcut for that long addition. This leads, as we’ll see later, to the idea of the “weighted mean”.)

Themodeis the type of average you want to know in this situation. The mode is the number thatoccurs most frequently. In the example for median, 6 would be the mode because it occurs twice, while the other numbers each occur once. In our employee example, the mode is $100 because that number occurs ten times, which is more than any other number occurs.

Here is a similar question about the need for different averages, giving a researcher’s perspective on the various statistics:

Using Mean, Mode, and Range

I’ll close with this question, from 2010:

Remember the Mean -- and the Median, and the Mode, and the Range, and the Outlier I need help with mean, median, mode, range, and outlier.I just can't remember which one's which-- especially the first four (I kinda already know "outlier"). All of them are rather hard. I don't have the first idea of how to do them, so I can't show any work. Andsince most of them begin with the letter "m," it gets confusing. My math teacher hasn't give us any memory strategies (or if she has, they weren't memorable). Also, I have a b-i-i-i-g-g-g AIMS test tomorrow about data analysis. I want to do well, especially since math is my worst subject. I'm sure you get that a lot. So if you could respond today, that would be awesome. But p-p-l-l-e-e-a-a-s-s-e, by tomorrow. Thank you, Dr. Math!

I answered this one:

Hi, Riley. The main way to learn which word has which meaning is just touse them a lot. Do you have some trick to remember which of your friends has which name? Probably not (unless they are twins and you don't know them well as individuals). Once you've spent enough time with someone, you know who they are! The same is true with vocabulary like this. Having said that, let's see if we canmake the meanings more memorable, so you can get to know them as individuals, keeping their meaning in mind as you use them. Anoutlieris a number that doesn't fit in with others in its group -- sort of aloner. It "lies outside" of the main group. Therangeof a set of numbers is how far they are spread out -- just as therange of a cell phoneis how far you can go and still talk, or therange of an animalis how far it travels.

Everyday usages are often not as far from the technical usage as you think, when you look at them right!

Themeanof a set of numbers is just another word for theaverage-- add them up and divide by how many. Themedianis the "middle" -- the number in themiddlewhen they are lined up in order. If there are two middle numbers, calculate the median by taking their average (mean). Themodeis the "most common." You may hear the word "mode" as in "a la mode" (originally meaning "fashionable") or "modish" (also meaning "stylish"). These mean that something is, in a sense, popular. The mode is themost "popular"value. One last tip: of the three terms that start with the letter "m," "median" is the one that sounds most like "middle."

We could also say that “mean” is what we usually *mean* by “average”; “median” is *in the middle* alphabetically; and “mode” is … the other one.

A question from the end of August led a student and a Math Doctor to an extra challenge, by way of an apparent typo in the problem. We particularly enjoy working with students who are willing to take on extra work in order to learn more than they need to!

Here is the question:

I have been given the following problem.

Find the sum of

1.2.3 + 2.4.6 + 3.6.12 + ……. to n terms

Here is how I tried,

The rth term is (r(r+1)/2).(2r).(3.2^(r-1)).

I could not do the summation because of the last geometric sequence term. Request you to help me.

One more important thing is that instead of giving the answer, the book has given the formula for the nth term of the sum. It is (3n^(2).(n+1)^(2))/2.

I checked it for the given 3 terms and it works.

Regards,

Rahul.

Rahul is carefully using *r* as the index of terms in the series, as distinct from *n* which is the number of terms in the series; others might have used *i* or *k*.

Doctor Rick replied:

Hi, Rahul. You write:

Find sum of 1.2.3 + 2.4.6 + 3.6.12 + ……. to n terms

Here is how I tried, The rth term is (r(r+1)/2).(2r).(3.2^(r-1))I think you may have made an error here by writing the formula for the

sumof the n first factors rather than the first factor itself. We can’t becertainabout the pattern of the terms from just three terms (or, really, any finite number of terms!), but it seemslikelythat the first factor of the rth term, 1, 2, 3, …, is just r. Instead you wrote the sum 1 + 2 + 3 + … + r, which is r(r+1)/2.For the other two factors of each term, we have the sequences 2, 4, 6, … and 1, 2, 4, …, which fit the formulas 2r and 2

^{r – 1}. Thus, again, though we can’t be sure this is the intention, it’s a reasonable assumption given that we areforcedto make an assumption!

So a corrected formula for the *r*th term of the series, as it appears to be, is $$a_r = (r)(2r)(3\cdot2^{r-1}) = 3r^2 2^{r}$$

One more important thing is that instead of giving the answer, the book has given the formula for the nth term of the sum. It is (3n^(2).(n+1)^(2))/2. I checked it for the given 3 terms and it works.The first point to notice here is that the problem is looking for the formula for the

sum of the first n terms of the series. This is often denoted as S_{n}, the nth term of the sequence of partial sums of the series. If the book’s answer calls it “the nth term of the sum”, it must mean the nth term of the sequence of partial sums. It surelydoesn’tmean the nth term of theseries.

So the answer the book gives is presumably just what it asks for, though the wording taken literally would mean the *n*th term \(a_n\) of the series (summation), which would be our formula above. The book is saying that $$S_n = \frac{3n^2(n+1)^2}{2}$$ But is that right?

My next observation is that when I check this formula “for the given 3 terms”, I do

notfind that it works. We havea

_{1}= 1*2*3 = 6 … S1 =6a

_{2}= 2*4*6 = 48 … S2 = 6 + 48 =54a

_{3}= 3*6*12 = 216 … S3 = 54 + 216 =270while from the book’s formula I get

S

_{n}= (3n^{2})(n+1)^{2}/2S

_{1}= (3*1^{2})(1+1)^{2}/2 =6S

_{2}= (3*2^{2})(2+1)^{2}/2 =54S

_{3}= (3*3^{2})(3+1)^{2}/2 =216It checks out for the first two partial sums, but not for the third. Did you get different results?

So either the *terms* shown in the problem are wrong, or the *formula* given for the sum is wrong! But look at the book’s formula for the sum: Does it look right for a sum of terms as given by our formula?

Now, the fact that

your formula for the rth termof the series has a geometric component (quite reasonably, since the sequence 1, 2, 4, … is the start of a geometric sequence), whilethe book’s formuladoes not look geometric at all, suggests that either we need to assume someotherpattern thatalsostarts 1, 2, 4, …, or that the series given in the problem has a typographical error in it! I think the latter is likely, since as I showed, the book’s answer does not work for the series given.As it turns out, when I first saw the problem I misread it as

1·2·3 + 2·4·6 + 3·6·

9+ …I suppose I did that because I was subconsciously making it a much easier problem — and in fact, the sum of

thisseries is given by the book’s answer. Thus my theory is that there is a typo in the problem, and that it was intended to have 9 instead of 12.What do you think of this?

Changing one number in the problem makes the book’s answer correct, so that seems a likely fix.

Rahul answered,

Dear Sir,

There seems to be typing error in the book about the answer or the question itself.

The problem is as follows,

Find the sum

1*2*3 + 2*4*6 + 3*6*12 + ……. up to n terms.

The answer given in the book also is not right. As you pointed out my mistake also, it is neither the formula for the sum up to n terms nor the formula for nth term.

Recognizing that there is an error, Rahul now showed in detail how to obtain the formula for the *r*th term in the series *as given*, correcting his original attempt (but not, yet, the problem itself):

In fact my work should have been as follows,

The first factor in the first term is 1, and the same in the second term is 2, and the same in the third term is 3, so I thought it is an arithmetic sequence with a=1, d=1, so I used the formula for its nth term as the general formula for the first factor of the nth (or the rth term before summing all the n terms).

It was 1+(r-1)*1 =

rThe second factor in the first term is 2, and the same in the second term is 4, and the same in the third term is 6, so I thought it is an arithmetic sequence with a=2, d=2, so I used the formula for its nth term as the general formula for the first factor of the nth (or the rth term before summing all the n terms).

It was 2+(r-1)*2 =

2r.The third factors were difficult to understand at first sight as it were not clearly arithmetic progression.

So I used the formula for Geometric progression and found the 3rd factor of the rth term as aq^(r-1) where q = 6/3 = 12/6=2. So it was

3*2^(r-1).

The product of these three factors is the formula I showed above for the *r*th term, \(a_r = (r)(2r)(3\cdot2^{r-1}) = 3r^2 2^{r}\). But this is evidently not the series the author intended.

So let’s assume the problem is really this:

Find the sum \(1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+\dots\) up to *n* terms.

The *r*th term is then \((r)(2r)(3r) = 6r^3\), which is much simpler.

Rahul continued:

So finally,

There is typing error in the problem, if the answer works for this correct problem instead 1·2·3 + 2·4·6 + 3·6·9 + …

And yes the answer is for this last sum itself.

That is Sum (r goes from 1 to n) of 6*r^(3) = (3*n^(2)*(n+1)^(2))/2.

Thank you so much for the help and revealing what should have been the correct problem itself.

Rahul hasn’t shown the work for obtaining this formula; I’m not sure what method he is expected to use; but the sum of cubes is a standard formula one can memorize or look up, which may be all that he did: $$\sum_{r=1}^n r^3 = \frac{n^2(n+1)^2}{4}$$ Memorably, this is the square of the arithmetic series: $$\sum_{r=1}^n r^3 = \left(\sum_{r=1}^n r\right)^2$$

For explanations of ways to obtain this and similar formulas (mostly using sums of other powers as examples), see:

Summing Integers to the Fourth Power Finding Sum Formula using Sequences of Differences Induction Problem Sum of Consecutive Squares Sigma Notation

Doctor Rick said,

Good, I believe you are satisfied with the idea that there was an error in the problem or the solution. It can be fun to take on the challenge of figuring out what an erroneous problem was likely

to be, and this helps convince us that it isn’tintendedourfault that we don’t get the “correct” answer!If you would still like to solve the problem as stated, even though it doesn’t seem to be the intended problem and is significantly more difficult, I believe the method(s) given in this item from the Ask Dr. Math Archive will be of help:

Here is the question and the initial answer from the page:

Finding the Sum of Arithmetico-Geometric Series Find the sum of the infinite series 1/7 + 4/(7^2) + 9/(7^3) + 16/(7^4) +... I would also like to know if there is a general rule to find the sum of (n^2/p^n) for n = 1 to infinity. The series looks like a convergent sequence. I wonder if there is a general rule to break a convergent sequence like above into one or more convergent sequences?

Though this is an infinite series, not a finite sum like our problem, then terms are quite similar to ours.

Doctor Vogler’s answer:

Hi Sudheer, Thanks for writing to Dr. Math. Have you ever summed an "arithmetico-geometric series"? That is, one of the form k S = sum a * n * r^n ? n=1 Some people do this by taking derivatives, but I won't assume that you know calculus, and I'll do it the algebraic way: Subtract r*S. k k S - r*S = sum a * n * r^n - sum a * n * r^(n+1) n=1 n=1 k k+1 S - r*S = sum a * n * r^n - sum a * (n-1) * r^n n=1 n=2 k+1 S - r*S = a * 1 * r^1 + [ sum a * r^n ] - a * k * r^(k+1) n=2 and then the middle series is a geometric series, and I assume that you can sum a series like that. This gives you a formula for S - r*S, and then you just divide both sides by 1-r to get S. Well, you have a series of the form k S = sum a * n^2 * r^n n=1 In fact, you can sum it in exactly the same way: Subtract r*S, regroup the sums, divide by 1-r, and you will get a formula for S which has an arithmetico-geometric series on the right, the type that I already solved. Use the formula for that kind of series to get the formula for your kind of series.

Note that this first defines an arithmetico-geometric series, whose terms are the product of an arithmetic sequence and a geometric sequence, and shows the technique for solving it; then he states that the problem posed is *not* that, but can be solved *similarly*.

Also, don’t let it confuse you that he uses *r* and *n* differently than in our problem; here *n* is the index in the series (our *r*), *k* is the number of terms (our *n*), and *r* is the common ratio (our 2).

Rahul answered, not lingering on the apparently correct problem, but leaping to the challenge of the problem as written:

Thanks for the reply.

I am taking a look at the Arithmetico-Geometric Series and definitely try to solve the as stated problem.

The book also mentions about the Arithmetico-Geometric series.

I am not able to show that it is arithmetico-geometric series because the first two factors of each term do not multiply to give an arithmetic sequence.

I mean it comes out to me as 2*3 + 8*6 + 18*12 + …. to n terms.

And here the factors 2, 8, 18 do not give me an arithmetic progression so as to use the formula for a general arithmetico-geometric series like, (a+(n-1)*d)*(g*r^(n-1)).

I have the first factor 3*2^(n-1) but not the first one.

Request you to help.

Doctor Rick corrected the misunderstanding,

The title of the link I provided is somewhat misleading, because it is not

onlyabout arithmetico-geometric series. The initial question is about a series more like ours — the term-by-term product of aquadraticseries and a geometric series. Doctor Vogler first reviews summation methods for an arithmetico-geometric series, and then shows how to change our kind of series into that:Well, you have a series of the form k S = sum a * n^2 * r^n n=1 In fact, you can sum it in exactly the same way: Subtract r*S, regroup the sums, divide by 1-r, and you will get a formula for S which has an arithmetico-geometric series on the right, the type that I already solved. Use the formula for that kind of series to get the formula for your kind of series.Can you follow this guidance? I haven’t tried it myself yet.

Rahul responded, having first tried to replicate Doctor Vogler’s work before moving to his own:

Dear sir,

I tried to repeat Dr. Vogler’s work in that reply but I am getting the different answer after simplification as follows

a1r + ( sum from n = 2 to k (instead of k+1) of ar^(n) ) – akr^(k+1).

Can you please help?

Request you to help only for this step. Then again I will move forward

Doctor Rick answered,

I believe you’re referring to the derivation of the last line in the following part of Dr. Vogler’s first response:

k k S - r*S = sum a * n * r^n - sum a * n * r^(n+1) n=1 n=1 k k+1 S - r*S = sum a * n * r^n - sum a * (n-1) * r^n n=1 n=2 k+1 S - r*S = a * 1 * r^1 + [ sum a * r^n ] - a * k * r^(k+1) n=2I didn’t think you would need to work on this part, since you indicated that your book “mentions” arithmetico-geometric series. I guess it doesn’t fully teach them, or else you haven’t studied that part. (I have never studied this topic myself.)

In going from the second line to the third, we want to combine the two sums; but in order to do that, we must first pull out terms that they do not have in common — the n=1 term in the first, namely a*1*r^1 = ar, and the n=k+1 term in the second, namely a*r^(k+1). This leaves summations from n = 2 to k, and subtracting term by term, we obtain sum[n=2 to k] (a*n*r^n – a(n-1)r^n), or sum[n=2 to

k] a*r^n.So I agree with you, there is an error in Dr. Vogler’s work. Unfortunately, we can no longer request corrections in the

Ask Dr. MathArchive.

To correct Doctor Vogler’s page, we would need to change the upper limit of the summation in the third line from *k* + 1 to *k*. But that’s a minor typo.

In any case, I would not use what Dr. Vogler wrote at all — I’m not interested in obtaining a general formula. (If I worked regularly with such series, I might want that!) All I’d do is to

apply the method to the specific series of interest.What arithmetico-geometric series will we want to sum? As Dr. Vogler suggests, we can apply the same

ideato our quadratic-geometric series. Again, there is no need (as far as I am concerned) for generality. We want to sum the seriesΣ

_{[k=1 to n]}(6k^{2}2^{k-1}) = 3 Σ_{[k=1 to n]}k^{2}2^{k}We will just look at the last sum, ignoring the multiplication by 3. Call it S:

S = Σ

_{[k=1 to n]}k^{2}2^{k}= 1^{2}2^{1}+ Σ_{[k=2 to n]}k^{2}2^{k}= 2 + Σ_{[k=2 to n]}k^{2}2^{k}2S = Σ

_{[k=1 to n]}k^{2}2·2^{k}= Σ_{[k=1 to n]}k^{2}2^{k+1}= Σ_{[k=2 to n+1]}(k–1)^{2}2^{k}= Σ

_{[k=2 to n]}(k–1)^{2}2^{k}+ n^{2}2^{n+1}2S – S = n

^{2}2^{n+1}– 2 + Σ_{[k=2 to n]}((k–1)^{2}– k^{2}) 2^{k}S = n

^{2}2^{n+1}– 2 + Σ_{[k=2 to n]}(1 – 2k) 2^{k}That’s as far as I’ll take it (and I don’t guarantee I made no mistakes myself). The sum in the last expression for S is the difference of a geometric series and twice the arithmetico-geometric series Dr. Vogler discussed.

He has modified the procedure a bit, for example subtracting S from 2S rather than vice verse for convenience. I made a couple small corrections here, so if there is an error, it may be due to either of us.

Rahul answered,

Dear Sir,

Thanks for making me work hard.After a lot of work I finally found the answer, here it is,

The sum S = 3*(2^(n+1)-4-2*(n*2^(n+1)-2(2^(n)-1))+4-2+n^(2)*2^(n+1)).

This time I am sure that I have checked it also correctly.

It was really a difficult fun.I am very happy that I could learn lot of stuff unknown to me earlier.

Thank you very much for the guidance.

I like that phrase, “a difficult fun”. I tell my students that my definition of “fun” is “challenging”, so they’re ready when I end a session with, “Wasn’t that fun?” I’m not going to try to fill in all the gaps; that would take away some of the fun for you, the reader.

Let’s check that \(S_3 = 270\), as we found from the three given terms:

$$S_3 = 3\left(2^{3+1}-4-2\left(3\cdot2^{3+1}-2\left(2^{3}-1\right)\right)+4-2+3^22^{3+1}\right) \\= 3\left(16-4-2\left(3\cdot16-2\left(8-1\right)\right)+4-2+9\cdot16\right) \\= 3\left(16-4-2\left(3\cdot16-14\right)+4-2+9\cdot16\right) \\= 3\left(16-4-68+4-2+144\right) = 3(90) = 270$$

Doctor Rick replied,

I’m glad you enjoyed the challenge! You have learned more than I did, because you did more of the work.

I checked your formula by putting it into a spreadsheet, and it agrees with the directly calculated sum as far as Excel can calculate exactly, which is n = 22. This strongly suggests that you solved the problem correctly. We can simplify it:

S[n] = 6(n^2 – 2n + 3)2^n – 18

Simplification isn’t essential (Rahul just wanted to get to an answer he could check), but it does make it look more like a final answer:

$$1\cdot 2\cdot 3+2\cdot 4\cdot 6+3\cdot 6\cdot 12+\dots \text{ (n terms) }= \sum_{k=1}^n 3k^2 2^{k} = 6\left(n^2 – 2n + 3\right)2^n – 18$$

]]>The first question is from 1998:

Spaces Formed by Intersecting Planes Do you know of a proof that would be used to showhow many subspaces can be formed by the intersecting of five planes in space?This should be the largest number, of course. Pascal's triangle and other patterns can lead to the conclusion that there is 1 space divided by 0 planes, 2 spaces divided by 1 plane, 4 spaces divided by 2 planes, 8 spaces divided by 3, 15 spaces divided by 4, and 26 spaces divided by 26 planes. But is there a proof?

The numbers given here are correct, as we’ll see; it isn’t clear how Kirste is using Pascal’s triangle (the numbers are not obviously present there), so we can’t be sure how close she came to a proof.

Doctor Tom answered, working step by step up to this three-dimensional problem:

Hi Kirste, There's a nice way to look at this bylooking at the problem in lower dimensions. In the easiest case,1 dimension, suppose you havea lineand you askhow many regions it can be divided into by n points. That's easy: each point divides one of the existing segments into 2, so each new point adds 1, so the total is n+1. (When n is zero, there's one line; adding a point makes 2 half-lines, and so on.)

So in the one-dimensional case, the formula is $$r(n) = n+1$$

We start with 1 region, and add 1 more for each point (new points are in blue here):

Now look at2 dimensions - a plane divided up by n lines. By messing around, you can see that for n = 0, 1, 2, 3, the answers are: 1, 2, 4, 7 regions, right? Well, suppose you've worked it out for some number n of lines, and you add the next line. In general, it will hit all n lines, so if you look at the intersections of the old lines with your new one, it gets hit n times, right? So the new line is divided into n+1 segments. (We just worked this out in the previous paragraph). So each of those n+1 segments will divide an existing region into two regions, so there will be n+1 new regions created, so you can work out the number of 2-D regions you get with n lines: 1 + 1 + 2 + ... + (n+1) (The initial "1" is because there's already one region when you start.) So the values from the formula above are: lines regions 0 1 = 1 1 1+1 = 2 2 1+1+2 = 4 3 1+1+2+3 = 7 4 1+1+2+3+4 = 11 5 1+1+2+3+4+5 = 16 ...

This is a quick version of what we saw two weeks ago about cutting a circle with *n* lines, but now we’re seeing it as an extension of a pattern going from one dimension to the next. Each (*n* + 1st) line we draw cuts each previous line, so that it is divided by *n* points, and therefore is divided into *n* + 1 parts, and adds *n* + 1 regions in the plane (new lines and points are in blue here):

For example, that third line adds 3 regions, one for each part into which it is itself divided. when we get to *n* lines, we have started with 1 region, then added *k* regions for the *k*th line, for a total of \(1 + 1 + 2 + 3 + \dots + n\). We’ll turn this into a formula soon.

Now go tothree dimensions. Assume you know the answer for n planes, and you want the answer for n+1. Well, the (n+1)st plane will hit all the n planes in one line each, so that plane is hacked into the number of regions that n lines will create, which we just worked out as the sum above. Each of those plane regions will divide a volumetric region into 2 pieces, so the answers for 3 dimensions are: planes regions 0 1 = 1 1 1+1 = 2 2 1+1+2 = 4 3 1+1+2+4 = 8 4 1+1+2+4+7 = 15 5 1+1+2+4+7+11 = 26 6 1+1+2+4+7+11+16 = 42 ...

Each (*n* + 1st) plane cuts each previous plane in a line, so that it is divided by *n* lines, and therefore is divided into the number of parts from the previous sequence, thus adding that many new regions to space (new planes and the lines that divide them are in blue here):

Summarizing:

Let me list all the results above: number of points, lines, planes: 0 1 2 3 4 5 6 7 dim -------------------------- 1 1 2 3 4 5 6 7 8 ... 2 1 2 4 7 11 16 22 29 ... 3 1 2 4 8 15 26 42 64 ... The first row is obvious - to get any other number in the chart,add together the number to its left to the number above it.

That is, for example, 3 + 4 = 7, and 16 + 26 = 42:

You can actually find a formula for it if you like. For dimension 1, it's linear: n+1. For dimension 2, it'll be quadratic, and for dimension 3, cubic. You can find a good reference for finding the formula for dimension 2 at: http://mathforum.org/dr.math/problems/regimbald3.9.98.html And by the way, this works fine for hyperspace of still higher dimensions.

The page he referred to is much like one that we saw two weeks ago, for lines in a plane rather than a circle. It turns a difference equation into a sum. Let’s try doing that here.

We saw first that the number of regions in a line cut by *n* points is $$r(n) = n+1$$ This is the first row in the table.

Then we found that the number of regions in a plane cut by *n* lines starts with 1, then increases sequentially by the number of regions in each new line, cut by as many points as there are lines already; so each number in the second row of the table is 1 plus the sum of numbers to the left in the row above (that is, from 0 through *n* – 1). So $$r(n) = 1 + \sum_{i=0}^{n-1}(n+1) = 1 + \frac{n(n+1)}{2} = \frac{n^2+n+2}{2}$$ That’s the formula for the second row.

Our goal is the third row. Again, we found that the number of regions in space cut by *n* planes starts with 1, then increases sequentially by the number of regions in each new plane, cut by as many lines as there are planes already; so each number in the third row of the table is 1 plus the sum of numbers to the left in the row above (that is, from 0 through *n* – 1). So (using a known formula for the sum of sequential squares) $$r(n) = 1 + \sum_{i=0}^{n-1}\frac{n^2+n+2}{2} = 1+\frac{n(n-1)(2n-1)}{12} + \frac{n(n-1)}{4} + n = \frac{n^3+5n+6}{6}$$

As we did two weeks ago, we could instead have obtained this formula by seeing the table as a difference table, upside-down, which tells us that the formula will be a cubic polynomial, and solving four equations for the four coefficients.

A similar question was asked again in 2001:

Planes Intersecting Space A plane divides space into at most two parts. Two planes divide space into at most four parts. Three planes divide space into at most eight parts. Four planes divide space into at most sixteen parts. Can we say thatn planes divide space into at most 2^n parts?Is there a proof of this idea?

Gafur, like many, has fallen into the now-familiar trap of seeing a false pattern; the last number, 16, is wrong.

Doctor Jubal answered this time, starting by showing why 16 doesn’t work:

Hi Gafur, Thanks for writing to Dr. Math. Actually,4 planes cannot divide space into 16 parts, but only 15. Let's say you have three planes that divide space into 8 parts. They intersect at a single point. Let's say you take a fourth plane that is not parallel to any of the other three, but also passes through the point. It will intersect each of the other three planes in a line, and these three lines will intersect at a single point.

Here are the three planes:

Here we see that the fourth plane, cut by three lines, misses two of the 8 regions, so it adds only 6:

That isn’t so easy to see, so we can make it easier:

You can represent this on a sheet of paper by drawingthree lines that intersect at a single point. The sheet of paper is the fourth plane. The three lines are the lines where it intersects the other three. You will see that the three lines divide the plane into 6 regions. The first three planes divided space into eight regions. Of these, six of them are divided by the fourth one, if the fourth one passes through the point of intersection of the the other three planes.

Here is our fourth plane:

With three lines, we’ve (so far) divided this plane into 6 regions; so it must pass through 6 regions of space, adding only 6 to our previous 8 for a total of 14 regions:

Of the other two regions, one is entirely "above" the sheet of paper, and the other is "below" it. Since the fourth plane doesn't have to pass through the point of intersection of the other three planes,it is possible to divide either one of these regions in addition to the original six. Make a new drawing on your sheet of paper. This time, draw three non-parallel lines thatdon't intersect in a common point. Instead, each pair of lines intersects each other at one of three different points. You will see they divide the piece of paper into seven regions.

This is the case when the fourth plane does not pass through the point of intersection of the other three. It is not parallel to any of the three original planes, so it intersects each of them in a line. Since it does not contain the point of intersection of the other three planes, these three lines do not intersect each other in a single point.

The planes look something like this:

But you don’t need to see this to know what’s happening!

You can see thatthis plane manages to divide seven out of the eight regionsthe first three planes divide space into. By not passing exactly through the point of intersection, the plane can divide one of the two regions it "missed" in the first case, while still dividing the other six.It is impossible for the plane to divide all eight regions, however, because it is impossible for the plane to pass to both "sides" of the point of intersection of the first three planes. As a result,four planes can divide space into at most 15 regions. Three planes divide space into eight regions, and a fourth plane can split all of these regions except one that it must miss because it can only pass to one side or another of the point of intersection of the first three.

How about a fifth plane?

In these four planes, there arefour points where three planes intersectto make a point. For each one of these points, a fifth plane must pass to one side or the other of it, and so it must miss four out of the 15 regions, and can only split 11 of the 15 into two. So five planes can only divide space into 15+11 = 26 regions.

The four points represent the 4 ways to choose 3 of the 4 planes: \(\displaystyle{{4}\choose{3}} = 4\).

In general, if you have n planes in space arranged that that no two planes are parallel and no two planes intersect each other in a line that is parallel to any other such line, those n planes will meet at n(n-1)(n-2)/6 points. (This formula comes about because you have n choices of the first plane, n-1 choices the second plane, and n-2 choices of the third plane, but the order of the planes doesn't matter, and there are six ways you could have chosen the same three planes.) At each of these points, the three planes that intersect at that point divide space into eight regions, only seven of which can be split by any single plane. So for each point of intersection of three planes, a new plane must "miss" one regions.

In other words, there are \(\displaystyle{{n}\choose{3}} = \frac{n!}{(n-3)!3!}\) triples of planes, each of which define an intersection point, and these are assumed to be distinct.

So if n planes divide space into at most r regions, the (n+1)st plane can only intersect r - n(n-1)(n-2)/6 of them, so n+1 planes can divide space into at most 2r - n(n-1)(n-2)/6 regions. Thus, 4 planes can divide space into 15 regions, 5 planes can divide space into 26 regions, 6 planes into 42 regions, 7 planes into 64 regions, 8 planes into 93 regions, 9 planes into 130 regions, 10 planes into 176 regions, and so on.

This is again a recursive formula; if we rewrite it to take *n* planes rather than *n* + 1, we get $$r(n) = 2r(n-1) – \frac{(n-1)(n-2)(n-3)}{6}$$ I won’t try to turn this into an explicit formula, since we’ve already seen one; but we can at least check it against previously verified numbers. For *n* = 7, we get $$r(7) = 2r(6) – \frac{(6)(5)(4)}{6}= 2(42)-20 = 64$$ which is correct.

Looking elsewhere, I have seen a number of interesting formulas for this. The most interesting, perhaps, is this: $$r(n) = {n\choose 0}+{n\choose 1}+{n\choose 2}+{n\choose 3}$$ This amounts to the number of spaces (1), plus the number of planes (*n*), plus the number of lines of intersection (pairs of planes), plus the number of points of intersection (triples of planes), counting up by dimension. And in fact, $${n\choose 0}+{n\choose 1}+{n\choose 2}+{n\choose 3} = 1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6} = \frac{n^3+5n+6}{6}$$ as we’ve seen before.

Now, we might well wonder whether this beautiful pattern is true in other dimensions.

We’ve already seen that the number of regions into which a line is divided by *n* points is *n* + 1; but that is equal to $${n\choose 0}+{n\choose 1} = 1+n$$

We’ve also seen that the number of regions into which a plane is divided by *n* lines is \(\displaystyle\frac{n^2+n+2}{2}\); but that is in fact equal to $${n\choose 0}+{n\choose 1}+{n\choose 2} = 1+n+\frac{n(n-1)}{2}$$

All we need is a direct way to get to these formulas. Can you find a way? I’ll leave that as an assignment for the reader. (It’s hidden in what we’ve already gone through!)

]]>Some questions we get, while small, raise interesting issues. In a question we got last month, there are several little issues pertaining to how the final answer should be chosen; as is often the case, it seems that a diligent student who cares about accuracy might be penalized. This has important implications for the design of tests. Let’s take a look.

Jing Zhu asked:

A question says,

What is the

smallest amount of moneythat a man needs to deposit annually for there to be an amount balance of over $15,000 at 6% interest after 12 years? (Round your answer to the nearest dollar.)According to the formula provided, the answer in decimal is $889.155. When 889 is plugged in the formula, the balance is

lessthan $15,000. When 890 is plugged in the formula, the balance is over $15,000. So, in this case, should the answer be889 according to the instruction statedinside the parentheses or890 according to the context?

To do a quick check before answering, I brought up Excel and used its FV (future value) formula to find the amount. With deposits of $889, the final amount comes to $14,997.38, which is indeed too little; it doesn’t answer the question! With deposits of $890, we end up with $15,014.25, which is enough. Jing Zhu is right (assuming payments are at the end of each year, which we’ll be looking at below).

We haven’t been told what formula was provided; that will be one issue (and is a reason we prefer that you show us the entire problem you are working on, include such context as formulas you use). The main issue is how to round.

I replied:

That’s a good question! I think it is a matter of interpretation.

I personally would want to round up to $890, as you suggest, because that is the smallest whole-dollar amount that will have the desired effect.If we take the problem literally, then we might say that

the answer is $889.16(where I rounded up to the nextpenny), but then round that as required to the nearest dollar to get $889. In this case, we are supposing that they don’t want an actual payment amount (which would be to the penny), but just want you toround for the sake of reporting your answer. This is particularly likely in a machine-grading context.I have seen book problems, and perhaps test problems, whose authors apparently just didn’t think that closely about the reality of the problem, but just automatically asked for a rounded answer.

If this is to be

graded by a human, I would state both answers with an explanation; it should get extra credit, for showing how well you are thinking! If it isgraded by a machine, then all you can do is what you think best. In that case, I would probably go with the literal answer, as machines don’t think carefully.And if there is a way to

ask the teacherabout it, I would do so.

One of my pet peeves is how computerized homework and testing can twist the way we are required to think. But even without computers, the same concerns surround multiple-choice questions or other formats where there is no way to show work. When work is shown, I often recommend explaining any assumptions you make (which may result in your solving the wrong problem, but conceivably getting full credit if your interpretation was reasonable and your work correct *for the problem you solved*).

Here the issue is at the other end of the problem: not an initial assumption about its meaning, but how you enter the answer at the end. When you aren’t sure how to round, I recommend prominently showing both the unrounded result and the result rounded as you think is requested or appropriate (or both, if they differ). But when all you can enter is a number, you are stuck in a dilemma.

(Similarly, when I teach, I often tell students clearly that even during a test, they are welcome to raise their hands and ask interpretation questions; this is especially important when I have students for whom English is a second language, or for whom the culture assumed in some “word problems” may be foreign. But it is good for the rest of the class, too! Some may not be familiar with some cultural detail I assume every American knows.)

Doctor Rick joined in with a slightly different perspective:

Hi, If I may add my own personal opinion, it is that the answer of

$890 is actually the answer to a different (and arguably better) question:What is the

smallest whole number of dollarsthat a man needs to deposit annually for there to be an amount balance of over $15,000 at 6% interest after 12 years?My view of the question as stated is that

the answer to the questionis $889.18, and then we follow thedirections for, by rounding it to $889.reportingour answerThere is another issue here, having to do with

details that are missing from the problem. Is the interestcompoundedannually? When are the deposits made — at thebeginning or endof each year? You say a formula was provided, so perhaps you didn’t have to know these details in order to choose a formula. Based on your answer, I supposethe formula providedwas some form of the first formula in the “Regular Deposits” section of this Ask Dr. Math FAQ:

This agrees with my thoughts, but says it more plainly: As a teacher, we would rather change the wording of the question so that the correct answer is clearly correct; as a student, we would take the wording literally and round after finding the truly correct answer.

And why does his wording result in a better question? This way, the rounding is an inherent part of the problem, not an afterthought tacked on for the sake of uniform answers. It takes rounding seriously, requiring thought about which way to round in order to fulfill the needs of the problem, which can be important in real life. And in so doing, it prevents the conflict we’ve seen in trying to answer it.

As for the formula, the FAQ says this:

Suppose you opened an account at a bank which was paying an annual interest rate of i (a fraction, equivalent to 100i%). You make a deposit of Mat the endof each of q equal time periods each year (including the end of the last period). The interest iscompounded once per period. Then the value P of the account at the end of n years is given by P = M([1+(i/q)]^{nq}-1)(q/i)

This formula, expressed more clearly, is $$P=M\left(\left[1+\left(\frac{i}{q}\right)\right]^{nq}-1\right)\frac{q}{i}$$ which can be rearranged slightly to $$P=\frac{Mq}{i}\left(\left[1+\left(\frac{i}{q}\right)\right]^{nq}-1\right)$$

For annual payments, \(q = 1\) (because payments are made once a year), and the formula becomes $$P=\frac{M}{i}\left(\left[1+i\right]^{n}-1\right)$$

This formula is derived from the formula for the sum of a geometric series, which is particularly easy to see in this case of annual payments. If payments are made at the end of each year, for *n* years, then the last payment earns no interest, adding only *M* to the final amount; the next-to-last payment grows for 1 year, adding \(M(1+i)\) to the total; and so on, until the first payment grows over \(n – 1\) years, contributing \(M(1+i)^{n-1}\). The total is therefore $$M+M(1+i)+…+M(1+i)^{n-1} = M\frac{\left(1+i\right)^{n}-1}{(1+i)-1}$$ which leads to our formula.

Jing Zhu answered us, providing the entire problem:

Thank you for your reply. I attached the original question for your reference.

This is an SAT prep question, and because of the word “over” (“over $15,000), I think we should use inequality to solve this problem. As you can see, the final answer turns out to be greater than 889.155. Therefore, I’d argue that the only correct answer should be 890.

The formula is the same as ours, where \(M\) is called \(P_0\), and \(i\) is called \(r\). But …

I replied,

Thanks. Interestingly, though the formula you were given is the same one we give in our FAQ that Dr. Rick referred to,

their description is wrong. This formula (for an “ordinary annuity”) applies when payments are made at theendof each period (year), not at thebeginningas they say (which is called an “annuity due”, and is mentioned after the main discussion in our FAQ). But we have to go by their formula, so we can ignore that discrepancy.

When payments are at the start of each period, each earns an extra period of interest, so the total is multiplied by \(\left(1+\frac{i}{q}\right)\), leading to the FAQ’s formula, $$P=M\left(\left[1+\left(\frac{i}{q}\right)\right]^{nq}-1\right)\left(1+\frac{q}{i}\right)$$ as shown in the FAQ.

The problem is valid if we change their word “beginning” to “end”. So the work will be the same whether we see this error or not. But if a student knew the correct formula, the problem would be still more confusing!

Here is the work to solve problem 38, as I would write it:

$$\frac{P_0}{0.06}\left(\left[1+0.06\right]^{12}-1\right)>15,000$$

$$P_0>\frac{(0.06)15,000}{\left[1.06\right]^{12}-1}$$

$$P_0>\frac{(0.06)15,000}{1.012196}=889.15544$$

So the smallest amount (in pennies) that satisfies this is found by rounding up to $889.16; and the smallest amount (in whole dollars) is found by rounding up to $890.

I continued:

Your use of an inequality, and your work itself, are good. We are left with the same dilemma we discussed: to

take it literallyand round thecorrectanswer of $889.16 to the (technicallyincorrect) answer, $889, just because they told you to; or toanswer Dr. Rick’s modified question, which makes more practical sense, and say $890.If I were you, I would answer $889, which seems to be what they are asking for, knowing that you are wiser than that. They said to find the correct answer, and

thento round that to the nearest dollar, so that is what you do.If this were given on an actual test, and it was a question on which you show work, you could

explain all this in writingand certainly get credit. If you could only give thenumerical answer, then I would enter $890.And it’s quite likely that the actual test would not have such a dilemma built in. (Sometimes there have been bad problems, and they have been discovered and accounted for.)

A similar point was made by teacher CTB in a comment to a post from last year, while I was writing this post:

Having myself also worked as an examiner (for A-level Maths – Stats module), it is unlikely that in an exam you would be penalised for any sensible decision you make to get to an answer. If you write down your assumption somewhere in your solution, so that the examiner can read it, you should get credit for your calculations.

When you can show work, you can be much more confident of being judged fairly, and therefore less anxious about how to answer.

We received this closing comment:

]]>Thank you for your detailed explanation. It was a pleasure to have had these discussions with you. Have a wonderful day! ^^

We can start with this question from Daniela in 1998:

Investigating Sequence Patterns This is an investigation we gave to our Year 8 maths students: You have a circle. You put 3 dots on the circumference and connect them with all possible straight lines, forming a certain number of regions (in the case of 3 dots, there are 4 regions). By putting an ever-increasing number of dots on the circumference, you have totry to find a pattern in the number of dots and the number of regions formed. Unfortunately, we could find no substantial pattern. Here is the table of results: dots 1 2 3 4 5 6 7 8 9 10 11 12 regions 1 2 4 8 16 30 57 88 163 230 386 456 At firstthere seemed to be a patterninvolving the Fibonacci sequence: dots regions 2 2 = 2*1 3 4 = 3*1+1 4 8 = 4*2 5 16 = 5*3+1 6 30 = 6*5 7 57 = 7*8+1 8 88 = here is where it started to go wrong We tried matching an exponential regression line to it, but it was not accurate. We even looked for patterns in the odd and even numbers of dots, thinking maybe it was a piecewise function; or patterns in the differences of the factors.... This investigation is being assessed, and any light you could shed on it would be greatly appreciated!

Here are the first few cases, showing 2, 4, 8, 16 regions for 2, 3, 4, and 5 *evenly spaced* points (colored to make them a little easier to count):

Doctor Schwa answered (already knowing the answer, as we’ll see, but going along with the discovery approach):

Wow, that's quite a lot of data. Did you draw a really big circle and try to count them? I would get tired way before this. Or did you have some other method of producing this table? If so,the method you used might be a good clue for uncovering the pattern. Also, not knowing where you got these big numbers makes me wonder ifthere might be some small mistake, and as you already know, even if you're just off by 1 you could hide a beautiful pattern. Noticing the Fibonacci sequence connection is a very nice start. I wouldn't have noticed that pattern at all, though Fibonacci does make some sense here because the pattern you unearthed is really subtle and hard to see. I'll do my best to help, but you've already used such a great set of problem solving strategies I don't know if I'll be able to do much better.

Although the Fibonacci connection is interesting, we’ll see that it is wrong because the numbers aren’t quite right!

Finding a pattern in the *way* you count, rather than just in the *numbers* you get, is a key idea also discussed in the post Building Patterns and Sequences.

Here are your data one more time to refresh my memory: dots 1 2 3 4 5 6 7 8 9 10 11 12 regions 1 2 4 8 16 30 57 88 163 230 386 456 I hope you were careful in drawing your dots not to make any special patterns with them (e.g. aregular hexagon, where several lines all pass through the point in the middle, will have fewer regions than if you used a nonregular hexagon).

Here are the next two cases, for *n* = 6 (twice) and *n* = 7. Observe that the first is a regular hexagon, and three diagonals meet in the center; in order to get the maximum number of regions, I’ve moved one point aside in the second picture, adding one region:

So for *n* = 6, the correct number is not 30 but 31. (And if you were expecting 32 because of the apparent pattern, we’ll be discussing that later!) For *n* = 7, the regular heptagon works fine, showing 57 regions. (Go ahead and count them; I’ll wait!)

In working through this, I noticed (as I never had before) that the problem didn’t specify that we want the *maximum* number of regions; it’s possible that the intent was to use *regular polygons*. In that case, the problem that was posed is quite difficult, and in fact was only solved relatively recently! That problem is briefly discussed here (the page title is inaccurate, and you’ll have to add *n* to get the numbers in Daniela’s list above):

Intersections of Polygon Diagonals

But the problem Doctor Schwa is answering is one that can be solved by ordinary people, so we’ll stick with it.

Rather than start with numbers that we hope were counted right, it’s safer to focus on how you do the counting: the method is a clue, as he said:

The way I start working on a problem like this is todraw a picture of my own. It's clear why 1 dot does nothing new, and just leaves 1 circle. The second dot lets you draw one line, cutting the circle in half. The third dot lets you draw two lines, cutting two new regions. The fourth dot lets you draw two lines to adjacent dots, each of which cut one new region, and one line to the opposite dot, which crosses a line on the way there so it makes 2 new regions, and we get 1 + 1 + 2 = 4 new regions all told.

Each time we add an *n*th point, we add *n* – 1 new lines (in blue), and one new region for each segment of those lines:

For example, the last figure shows adding the 4th point, which adds 3 lines, which consist of a total of 4 segments; each segment split an existing region, so we have 4 new regions, raising the total from 4 to 8.

So then I start wondering if I can make a pattern out of that. I don't see it yet, so I scribble a couple more cases on my picture. The fifth dot connects to two adjacent dots again, making one new region each, and now there are two "opposite" dots, and I have to cross two lines on the way there, so they make a total of three new regions each, for 1 + 1 + 3 + 3 = 8 new regions.

So now I guess the pattern: when I add the sixth dot, I'll get 1 + 1 for the adjacent two dots I connect to, 4 + 4 for connecting to the two dots next to them, and then 5 for connecting to the opposite vertex. That means I added 15, so I get 31 regions for 6. Maybe you did the problem with a regular hexagon? Or some other kind of shape where more than two lines crossed at one point? Or you missed counting one region? Or maybe I made a mistake? If I did, let me know.

Supposing I'm right so far, the pattern does get pretty complicated. I always add 1 + 1 for connecting to the two adjacent vertices, then (n-2) + (n-2) for the two next to that but it's tricky to count how many crossings I make going to the next vertices. The way I think of it is thatthe number of new regions created is the number of connections I cross, plus 1. Each vertex to the left of the line I draw has to connect to each vertex to the right of the new line I draw. So when I draw that 7th dot, I get two adjacent dots giving 1 + 1, the two next to that (splitting the other 5 dots into groups of 1 and 4) gives me (4+1) + (4+1), and the remaining two (splitting the other 5 dots into groups of 2 and 3) gives me (2*3 + 1) + (2*3 + 1) so all told I get 2 + 10 + 14 = 26 new regions. So my number for 7 dots would be 31 + 26 = 57. Looks like I'm back into agreement with you!

The agreement, as we saw above, is because with 7 points (a prime number) there are no multiple intersections among diagonals for the regular heptagon.

There may well be a simple pattern where I could find, say, the answer for 100 dots without having to carefully add up everything along the way, but if there is, I don't see it now. I could try a bunch of messy algebra to express this summation formula, but that doesn't seem as if it would be fun. And even if it turned out to be fun, it doesn't sound particularly promising. But perhaps this method of counting will let you generate a few more numbers in the list (and check, and correct some of the numbers you already have? Again, unless I'm misinterpreting the problem...) and maybe once you have a correct table you'll be able to find some pattern that I'm not seeing yet.

Actually, the pattern as described here, thought of from the right perspective, will lead to a neat solution that we’ll see in our last answer. Just to get you thinking: We start with one region; then for each new line we draw, we got one new region for each intersection we create, plus another region just for finishing the line …

But how about another way?

Last week, looking at regions formed by *n* chords, we saw that one method used the equality of second differences to determine that the formula was quadratic, and then we found a formula that fit algebraically. We can do the same here (now that we have used the pattern we found to confirm that we have accurate numbers):

I think themethod of finite differencesis very promising. In fact, just with the data we already have (if I'm right about the 31?): 1 2 4 8 16 31 57 1 2 4 8 15 26 1 2 4 7 11 where in each row I subtract the two numbers above it. I think a complete solution to the problem is not too far away at that point.

Let’s carry this out:

The next line (third differences) will be 1, 2, 3, 4, so the fourth differences are 1, 1, 1. This is enough to be reasonably confident that there is a fourth-degree formula for this sequence. So we can assume the formula is $$r = ax^4 + bx^3 + cx^2 + dx + e$$ and create five equations to solve for these five unknowns, by taking *x* = 1, 2, 3, 4, 5:

$$a(1)^4 + b(1)^3 + c(1)^2 + d(1) + e = 1$$

$$a(2)^4 + b(2)^3 + c(2)^2 + d(2) + e = 2$$

$$a(3)^4 + b(3)^3 + c(3)^2 + d(3) + e = 4$$

$$a(4)^4 + b(4)^3 + c(4)^2 + d(4) + e = 8$$

$$a(5)^4 + b(5)^3 + c(5)^2 + d(5) + e = 16$$

That is,

$$a + b + c + d + e = 1$$

$$16a + 8b + 4c + 2d + e = 2$$

$$81a + 27b + 9c + 3d + e = 4$$

$$256a + 64b + 16c + 4d + e = 8$$

$$625a + 125b + 25c + 5d + e = 16$$

I will not show the work, but I *have* actually done it, and the solution is \(a=\frac{1}{24}, b=-\frac{1}{4}, c=\frac{23}{24}, d=-\frac{3}{4}, e=1\), so that

$$r = \frac{n^4 – 6n^3 + 23n^2 – 18n + 24}{24}$$

A post on “finite differences” has been in the works for a long time, and will hopefully be coming soon! But for more information, you can search *Ask Dr. Math* for the phrase.

If you'd like to talk about this problem some more, please do send me an email. This kind of problem solving is a lot of fun. Or, if you'd like to see a more formal treatment, look at ...

Here is the question he referred to, which was actually posted only five days earlier in 1998:

Counting Regions Formed by Chords of a Circle Dear Doctor Math, This is an extension of a problem we have been dealing with in class. I would like your help in understanding it, if possible. Let n be the number of points on the circumference of a circle, and let r be themaximum number of non-overlapping regions formed by connecting each of the n points, with a line segment, to every other point on the circle, including its two neighboring points. Let n = 1, 2, 3, 4, 5, 6 .... Find a pattern relating n and r. Find an explicit or a recursive equation for r, and prove this formula. I would like any help you can offer with this question.It seemed straightforward, but then it just changes; and finding the formula, and being able to prove it, are not so simple. I would appreciate any help you can give me. Thank you. Lover of Math, Jay Shah

This one is very clearly stated. And, yes, the pattern starts out with 1, 2, 4, 8, 16, making you think the next will be 32, but it “just changes”.

Doctor Rob answered, first stating the formula that will be derived, in two forms:

The formula isr(n) = C(n,4) + C(n,2) + 1. The values for n = 1, 2, 3, ... are: 1, 2, 4, 8, 16, 31, 57, 99, 163, 256, .... Here, C(n,k) = n!/[k!*(n - k)!]. This expression for r can also be expressed in the form:r(n) = (n^4 - 6*n^3 + 23*n^2 - 18*n + 24)/24

That second form is just what we (well, I) found above. The first form is found most easily by a method I’ve been hinting at, which is the goal of this post. But for now, we have a method that starts with the pattern we already saw:

You can prove this by starting with n points, all connected, and adding one more point. Connecting it to its two neighbors adds 1 region each. Connecting it to their neighbors adds 1*(n - 2) + 1 regions each. Connecting it to their neighbors adds 2*(n - 3) + 1 regions each. Connecting it to their neighbors adds 3*(n - 4) + 1 each. Each line from the new point to an old vertex crosses k*(n - k - 1) lines connecting the k vertices on one side of the line to the remaining n - k - 1 vertices on the other side, andcrossing m lines results in m + 1 new regions. That means the number of regions added is: n-1 r(n + 1) - r(n) = Sum [k*(n - k - 1) + 1] k=0 n-1 = Sum [1 + n*k - k*(k + 1)] k=0 = n + n*n*(n - 1)/2 - n*(n - 1)*(n + 1)/3 = (n^3 - 3*n^2 + 8*n)/6 = n*(n + 1)*(n + 2)/6 - 2*n*(n + 1)/2 + 2*n

We are summing the numbers added by each new point, using a formula Doctor Schwa got close to but didn’t quite state. To carry out the summation, Doctor Rob uses known formulas for sums: \(\sum_{k=0}^{n}k = \frac{n(n+1)}{2}\) and \(\sum_{k=0}^{n}k(k+1) = \frac{n(n+1)(n+2)}{3}\).

Thus: n r(n) = r(0) + Sum [r(k) - r(k - 1)] k=1 n = 1 + Sum [(k - 1)*k*(k + 1)/6 - 2*(k - 1)*k/2 + 2*(k - 1)] k=1 = 1 + (n - 1)*n*(n + 1)*(n + 2)/24 - 2*(n - 1)*n*(n + 1)/6 + 2*(n - 1)*n/2 = 1 + n*(n - 1)*(n^2 + 3*n + 2 - 8*n - 8 + 24)/24 = 1 + n*(n - 1)*(n^2 - 5*n + 18)/24 = 1 + n*(n - 1)/2 + n*(n - 1)*(n - 2)*(n - 3)/24 = 1 + C(n,2) + C(n,4)

At the end, he is reversing the process, using well-known formulas for binomial coefficients C(n,k), which is also written as \(_nC_k\) or \({n}\choose{k}\). It would be more natural just to produce the quartic equation we found previously.

Frankly, I’ve included this, which is a huge amount of algebra that I don’t expect anyone to follow, just to make the next answer even more satisfying.

For a more direct way to the same formula, here is a question from 2003 (which was combined with an answer we looked at last time, to the other circle-region question!):

Circle Regions We drew a circle and then put two points on the line and joined the points. The circle is divided into two parts. If there are 4 points (not evenly spaced) and they are each joined to all the other points, the circle is divided up into 8 regions. There is aconjecturethat says that if the process is repeated, with more points along the edge of the circle, not evenly spaced, then a rule forthe maximum number of regionsis 2 to the power of (the number of points take one) r=2^(n-1) where ^ = to the power of. For 2,3,4,5,6 points, they follow the formula, but for 7 points, it doesn't; there are 57 regions, and if it followed the rule, there'd be 64. Am I not counting them properly? Is there a simple way of understanding how many regions there are, because there must be a constant formula, shouldn't there be?

The “conjecture”, which is false, is that because the first terms are 1, 2, 4, 8, 16, the general term is \(r_n = 2^{n-1}\). Sheila has seen that it doesn’t work. What does?

Doctor Anthony answered with his usual conciseness:

n points are distributed round the circumference of a circle and each point is joined to every other point by a chord of the circle.Assuming that no three chords intersect at a point inside the circle we require the number of regions into which the circle is divided. With no lines the circle has just one region. Now consider any collection of lines.If you draw a new lineacross the circle which does not cross any existing lines, then the effect is toincrease the number of regions by 1. In addition, every time a new linecrosses an existing lineinside the circle the number of regions is increased by 1 again.

This is the pattern I pointed out above: In looking at the number of regions added by a new point on the circle, we observed that we add one region each time we cross another line, and another when we finish the line. So each interior intersection corresponds to a region, as does each line drawn. We can count them all at once, rather than point by point. For example, looking again at the case of 5 points:

There was 1 region to start, plus one each for 10 lines, plus one each for 5 intersections, for a total of \(1+10+5 = 16\) regions.

Now, how can we calculate the numbers of lines and intersections?

So in any such arrangement number of regions = 1 + number of lines + number of interior intersections = 1 + C(n,2) + C(n,4) Note that thenumber of linesis the number of ways 2 points can be chosen from n points. Also, thenumber of interior intersectionsis the number of quadrilaterals that can be formed from n points, since each quadrilateral produces just 1 intersection where the diagonals of the quadrilateral intersect.

That is, there is one line for each pair of points, so we count combinations of *n* points taken 2 at a time, \(\displaystyle {{n}\choose{2}} = \frac{n(n-1)}{2}\). And there is one intersection for each set of four distinct points, because only one way of pairing them produces an intersection. For example, the set of four points shown here result in six different lines, but only one intersection:

So the number of intersections is the number of combinations of *n* points, taken 4 at a time, \(\displaystyle {{n}\choose{4}} = \frac{n(n-1)(n-2)(n-3)}{24}\).

And we can get the quartic formula from this: $$1+{{n}\choose{2}}+{{n}\choose{4}} = 1 + \frac{n(n-1)}{2} + \frac{n(n-1)(n-2)(n-3)}{24} = \\ \frac{24+12n^2-12n+n^4-6n^3+11n^2-6n}{24} = \frac{n^4-6n^3+23n^2-18n+24}{24}$$ which we have seen before.

Examples: n=4 Number of regions = 1 + C(4,2) + C(4,4) = 8 n=5 Number of regions = 1 + C(5,2) + C(5,4) = 16 n=6 " " = 1 + C(6,2) + C(6,4) = 31 n=7 " " = 1 + C(7,2) + C(7,4) = 57 Note that formula 2^(n-1) starts to go wrong at n=6

For an interesting question from 2001, where this week’s topic was confused with last week’s, read this:

Slicing Up a Circle]]>

Often the hardest part of solving a problem is interpreting what it means. Math is precise; human language can be ambiguous, and assumptions can be hidden. Today, we look at a multi-variable calculus problem that looked enough like a classic single-variable maximization problem to fool the reader into not taking it at face value – but as a result, we get to talk about two problems at once!

Here is the question, which came from Stefan last month:

Find the maximum volume of a box with its top removedwhich is supposed to be made out of12 m.^{2}of cardboardNow I’m not sure what exactly to do.

Should I write the function for volume and then find the extreme values?

12 = x*y, x = 12/y

V = (y – 2z)(x – 2z)z

V = (12/y – 2z)(y – 2z)z

Doing this and finding partial derivatives ∂V/∂x and ∂V/∂y I get x, y = √12 and z

_{1}= 0.57 and z_{2}= 1.73V

_{1}= 3 and V_{2}= 2.91042169×10^{-5}, I know that both of these aren’t correct.

Doctor Rick answered:

How do you know that both these solutions are incorrect? If an answer was provided, what was it?

The problem is not entirely clear. You seem to be reading it this way: “Find the maximum volume of a lidless box made by

starting with a rectangle of area 12 m, cutting squares out of the corners and folding up to make the sides.” This is a fairly common type of problem, but is not the only way the words of the problem could be interpreted.^{2}Solving the problem as you interpreted it, I get the same results you get, though when I find the exact answers (not just using a calculator and rounding), my answers are a little

3 for the greater (which must be the maximum volume) andmore than0 for the lesser.exactly

Here is Stefan’s interpretation of the problem, showing the variables he has in mind:

Let’s pause and do the work as Stefan presumably did it. He has found an expression for the volume *V* as a function of length *x*, width *y*, and height *z*; and he wants to maximize it. A problem is commonly given in this form with specific values for *x* and *y*, to be solved with only one variable. To find the maximum here, with two unknowns, we have to take the derivative of *V* with respect to one variable at a time, and set both to zero. If you are not familiar with this, a special symbol is used in place of the *d*‘s in the derivative, now called a ** partial derivative**, just to remind us that other variables are being treated as constants. Here are the function and its two derivatives:

$$V = \left(\frac{12}{y} – 2z\right)(y – 2z)z = 4z^3-2yz^2-24y^{-1}z^2+12z$$

$$\frac{\partial V}{\partial z} = 12z^2-4yz-48y^{-1}z+12$$

$$\frac{\partial V}{\partial y} = -2z^2+24y^{-2}z^2$$

Setting these equal to zero, we have a system of equations, $$\left\{\begin{matrix}12z^2-4yz-48y^{-1}z+12=0\\ -2z^2+24y^{-2}z^2=0\end{matrix}\right.$$

To solve this, we can rewrite the second equation as $$z^2\left(\frac{24}{y^2}-2\right) = 0$$ so that \(z=0\) or \(y=\sqrt{12}=2\sqrt{3}\). In the former case, the first equation becomes \(12=0\), so it is impossible (and the volume would be zero anyway). In the latter case, we find that \(x=\frac{12}{y}=\frac{12}{2\sqrt{3}}=2\sqrt{3}\), and the first equation becomes $$ 12z^2-16\sqrt{3}z+12=0$$ which factors as $$4(3z-\sqrt{3})(z-\sqrt{3})=0$$

So the two solutions are $$x=2\sqrt{3},y=2\sqrt{3},z=\sqrt{3}\Rightarrow \\V = \left(\frac{12}{y} – 2z\right)(y – 2z)z = \left(\frac{12}{2\sqrt{3}} – 2\sqrt{3}\right)(2\sqrt{3} – 2\sqrt{3})\sqrt{3} = 0$$ and $$x=2\sqrt{3},y=2\sqrt{3},z=\frac{\sqrt{3}}{3}\Rightarrow \\V = \left(\frac{12}{y} – 2z\right)(y – 2z)z = \left(\frac{12}{2\sqrt{3}} – \frac{2\sqrt{3}}{3}\right)\left(2\sqrt{3} – \frac{2\sqrt{3}}{3}\right)\frac{\sqrt{3}}{3} = \frac{16}{3\sqrt{3}} = 3.0792$$

The dimensions of the box are \(x-2z=, y-2z=\frac{4\sqrt{3}}{3}, z=\sqrt{3}\), so the ratio of dimensions is 4 : 4 : 3. But the numbers are such that, if the problem were intended to be taken this way, a different area than 12 probably would have been chosen!

Doctor Rick continued:

If this is not the intended answer, then we can try another interpretation. Perhaps the box is not made by starting with

onerectangle of area 12 m^{2}, but rather fromfiverectangles (four for the sides and one for the bottom) oftotalarea 12 m^{2}. (Alternatively you could start with three rectangles, one folded to make the bottom and a pair of opposite sides; it should come out the same.) I haven’t done the work for this yet.

In other words, if we take the problem literally, the box is made with 12 m^{2} of cardboard, with no scrap.

Stefan replied,

The solution is 4 m

^{2}, not cubed, so it might be an error in the book.Is there a reason to use the other methods or can I always cut the corners?

Doctor Rick answered, having now done the work:

With the second interpretation of the problem, I get a maximum volume of 4 m

^{3}, so I’d say this is the intended meaning of the problem and only the units were a typo.This isn’t about

method— any valid method for solving a problem should give the same result — but about understanding themeaningof the problem. As I said, it is not worded very clearly. In the absence of clarity, I suppose you chose to interpret it as being like other problems you have seen, but the problem itself does not give us strong reason to take it that way.Now that I am reasonably certain that I know what the problem is intended to be, would you like to try solving it? Again, this box is to be constructed from five rectangles of cardboard (for the front, back, left, right, and bottom of the box); the

total areaof the five rectangles is 12 m^{2}. What is the maximum volume (or capacity) of the box, and what are the dimensions of this maximum-volume box?

Here is the reinterpreted problem; observe that the meaning of the variables has changed:

Now we’ll have a considerably more complicated formula for the area that is 12 m^{2}, but the volume formula is far simpler.

Stefan wrote back,

So the area would be xy + 2xz + 2yz = 12, x(y + 2z) = 12 – 2yz, x = (12 – 2yz)/(y + 2z)

V = xyz, V = (12yz – 2y

^{2}z^{2})/(y + 2z)∂V/∂y = -2z

^{2}(y^{2 }+ 4yz – 12)/(y + 2z)^{2}∂V/∂z = -4y

^{2}(yz + z^{2 }– 3)/(y + 2z)^{2}After solving these I get that y = ±2z and from this that z = ±1, z has to be positive, so I have y

_{1}= 2 and y_{2}= -2, and y also has to be positive, from this I get x = 2 and so V = 2*2*1 = 4Is this how it’s meant to be done?

He again solved the area equation for *x*, and used that to write an expression for *V* as a function of *y* and *z*. The partial derivatives are correct. Setting them to zero, we can ignore the denominators, and the system we have to solve is $$\left\{\begin{matrix}y^2+4yz-12=0\\ z^2+yz-3=0\end{matrix}\right.$$

This looks hard to solve at first, but if we just subtract 4 times the second equation from the first, hoping to eliminate *yz*, everything goes right and we get $$y^2-4z^2=0$$ so that, indeed, \(y=\pm 2z\). Just as Stefan said, we can ignore the negative case, plug \(y=2z\) into the first equation to find that \(z=\pm 1\). So he’s got it. We now have the right solution to the right problem. (The first interpretation was just good exercise.)

There was a three day lapse before Doctor Rick could reply:

I was writing back to you when the power went out at my house due to a storm. It came back on last night, so I can finally send you my reply:

I did something similar but not identical, so I can’t compare your work with mine line by line, but you did get the same results I got. (The results are “nicer” than the results following the other interpretation.)

I chose to solve for the height, z, rather than for x. I did this because

the equations are symmetric under swapping of x for y, and this turned out to be helpful. My volume function wasV(x, y) = xy(12 – xy)/(2(x + y))

From ∂V/∂y = 0 I obtained the equation

y

^{2}+ 2xy – 12 = 0 (or y = 0)Solving this directly isn’t so helpful, but I can immediately write the equation for ∂V/∂x by switching x and y:

x

^{2}+ 2xy – 12 = 0 (or x = 0)Comparing the two equations, I see that x

^{2}= y^{2}, so y = ±x; and as in your work, I concluded that y = x because both must be positive. Putting y = x into the second equation, I got3x

^{2}– 12 = 0so x = y = 2 (since x and y are non-negative). Plugging these into the function V(x, y), we get V

_{max}= 4; if x = 0 or y = 0, then V(x, y) = 0, clearly the minimum.

By the way, solutions to this sort of problem are typically “nice” in some sense. Here, the optimal dimensions of the box are \(1\times2\times2\), and if we put two such boxes together we get a complete \(2\times2\times2\) cube.

]]>For the first problem, we can start with this question from 1998:

Regions, Chords, and Circles A given circle hasn chords.Each chord crosses every other chordbutno three meet at the same point. How many individual regions are in the circle? This is how much I've completed on this problem: I know that the xth line splits x regions, increasing the number of sections by x. I'm having trouble finding out the formula to solve the problem. Please help me!

Here is a simple example, with 3 chords dividing the circle into 7 regions:

We’ll be seeing different characterizations of the same problem, in which we are only asked for the **maximum** number of regions; that will turn out to imply this idea that each chord intersects all the others. For example, if my three chords intersected at one point, we would lose a region, and have only six:

On the other hand, if any chord does not intersect all the others, we again have fewer than 7 regions:

Doctor Jaffee replied, using an approach based on looking for a pattern in the numbers:

Hi Dana, Your observation about how the number of regions increases with each additional chord is right on target. We can use that to construct a table of values that can be helpful: number of chords 1 2 3 4 5 6 number of regions 2 4 7 11 16 22 You noticed that the number of regions increases by 2, then 3, then 4, etc., and that is one of the properties ofquadratic functions; that is, as the x number increases by 1, they number increases by a constantly increasing amount. In the example above each increase is 1 more than the previous increase.

Here are the first few cases, showing how each new chord *n* crosses each of the previous *n* – 1 chords, cutting off *n* new regions (one from each of the *n* regions it passed through), thereby adding 1, then 2, then 3, then 4 regions:

Doctor Jaffee has made a table of values, then used the fact that if the difference increases by the same amount at each step, the formula will have quadratic form. For example, looking at the perfect squares, 1, 4, 9, 16, we see that the differences from one term to the next are 3, 5, 7, and so on, always increasing by 2.

So, we can pick any three ordered pairs from the chart, substitute the values into the standard equation of aquadratic function(y = ax^2 + bx + c). We will then have three equations in three variables which we can solve. Substitute those numbers back into the standard equation and we'll have finished.

We’re going to solve for the three coefficients of the formula.

It looks like this: I would pick the first three pairs (1,2), (2,4), (3,7). Substitute them into the standard quadratic form and get 2 = a + b + c 4 = 4a + 2b + c 7 = 9a + 3b + c The solution to this system is a = 1/2, b = 1/2, and c = 1. So, substituting them into y = ax^2 + bx + c we get y = (1/2)x^2 + (1/2)x + 1 where x is the number of chords and y is the number of regions formed. You can then verify that if x = 4, y turns out to be 11 just as in the chart. If x = 5, y = 16, and so on.

One way to solve the system of equations is to first subtract each equation from the next one, leaving two equations in two unknowns: $$2=3a+b\\3=5a+b$$

Then we can subtract the first of these two from the second: $$1=2a$$ so that \(a=\frac{1}{2}\), and then, putting that into the first of the two equations, \(2=3\cdot\frac{1}{2}+b\), so \(b=\frac{1}{2}\); and then from the first original equation, \(2=\frac{1}{2}+\frac{1}{2}+c\), and \(c=1\).

In the terms I’ll be using, taking *n* = number of chords and *r* = number of regions, the formula is $$r = \frac{1}{2}n^2 + \frac{1}{2}n + 1 = \frac{n^2+n+2}{2}$$

A similar question was asked in 2001, and got two answers:

Dividing a Circle using Six Lines What is thelargest number of regionsinto which you can divide a circle with six lines?Is there a formula?How can you get this answer? I've tried drawing diagrams but I can't find a way to make sure they are correct. I can only get 14, but I know there are more.

Here the problem is phrased in terms of the maximum number of regions, rather than telling us each chord crosses all the others; and it asks about a specific number of chords, not a formula. But a formula can help Jordy know whether what he has drawn is the best he can do. And so will the concepts behind the formula.

Doctor Rob was the first to answer, providing the formula we just saw (without proof) and a way to actually draw the figure, which accords with our observations above:

Thanks for writing to Ask Dr. Math, Jordy. There is a formula. For n lines, you can get at most (n^2+n+2)/2 regions. When n = 6, this gives 22 regions.

Here is the benefit of having a formula; we can tell whether we have reached our goal in the drawing.

To maximize the number of regions, draw your lines like this: Start with one line. Label the two points where it meets the circle as P1 and Q1.

First chord:

Pick points P2 and Q2 such that they are encountered in the order P1,P2,Q1,Q2,P1 going around the circle clockwise. Connect P2 and Q2 to make the second line, and four regions.

Second chord:

If the new points were on the same side of the first chord, we wouldn’t get the maximum number of regions, so the order is important.

Between P2 and Q1 pick point P3. Between Q2 and P1 pick point Q3. Connect them with a line, but, if necessary, move Q3 a little to make sure that the line does not pass through any of the intersections of any previously drawn lines. That will give you seven regions.

Third chord:

Again, we have to cross both existing chords, so it has to start between the last P and the first Q, and end between the last Q and the first P.

Between P3 and Q1 pick point P4. Between Q3 and P1 pick point Q4. Connect P4 to Q4 with a line, but, if necessary, move Q4 a little to make sure that the line does not pass through any of the intersections of any previously drawn lines. That will give you 11 regions.

Fourth chord:

It’s getting harder to avoid intersections, as the regions are getting small.

Continue in this way until you have 6 lines and 22 regions.

Here is my result, after making many adjustments of existing points to make room:

Now he suggests a more orderly way to pick the points that will ensure that everything fits:

P1-P6 and Q1-Q6 can be chosen to be12 adjacent vertices of a regular n-goninscribed in the circle, if n > 12. It is easy to construct a regular 16-gon by starting with a diameter of the circle and bisecting the central angles until they are all equal to 360/16 = 22.5 degrees. That will locate 16 equally-spaced points around the circumference of the circle. Pick any 12 that are adjacent, and label them P1, P2, ..., P6, Q1, ..., Q6. Then connect P1Q1, P2Q2, ..., P6Q6. If you have done this carefully, you will be able to see and count the 22 regions.

This produces a more regularly spaced pattern that is easier to count:

Doctor Greenie, who probably had been writing as Doctor Rob wrote, stepped in with essentially the same approach, but this time leading to the formula:

Hi, Jordy - There is a formula - as I see another doctor here has already shown you. But there is a way you can get the answerwithout the formula, by logically analyzing what happens as you draw more and more lines. And, if you know the correct mathematical techniques, this logical analysis can lead you to aderivation of the formula. I have always found mathematics to be more enjoyable when I can see where a formula comes from, rather than having to regard it as some sort of mathematical magic. So here is how you can analyze this problem, at least far enough to come up with the answer to your problem. You are going to analyzehow and when the number of regions increasesas you draw more lines through the circle. You will find that, if you think of actually physically drawing each line,the number of regions increases by one at a time; each increase is the result of an existing region being cut into two separate regions. Start with a circle; before you draw any line through the circle, it is asingle region.

If you’ve followed the thinking as others showed it, you may skip the details below; but if you need any help being sure of the hows and whys, this should make it clear. If I drew pictures for this, they would be the same we’ve already seen:

Now, start drawing thefirst linethrough the circle, and consider at what point(s) you add another region to the diagram. When you are almost all the way from your starting point to the other side of the circle, there is still only one region; it is not until you reach the other side that you have divided an existing region (in this case, the entire circle) into two parts. So, when you draw the first line through the circle, there isone point at which one of the existing regions becomes two regions: when you reach the opposite side of the circle. So we have: 1st line drawn through circle ==> number of additional regions = 1 Next, start drawing thesecond linethrough the circle, and consider at what point(s) you add another region to the diagram. Of course, since you want to create the largest number of regions in the circle, you want your second line to intersect the first line inside the circle. So you start at some point on the circle, andwhen you reach the first line somewhere inside the circle you have added another regionby dividing an existing region into two parts. And then whenyou reach the other sideof the circle you have again added another region by dividing an existing region into two parts. So, when you draw the second line through the circle, there aretwo points at which one of the existing regions becomes two regions: when you reach the first line you drew, and when you reach the opposite side of the circle. So we have: 2nd line drawn through circle ==> number of additional regions = 2 Now, start drawing thethird linethrough the circle, and consider at what point(s) you add another region to the diagram. Of course, since you want to create the largest number of regions in the circle, you want this third line to intersect both of the first two lines at different points inside the circle (if your third line intersects the first two lines at their point of intersection, you don't get the maximum possible number of regions). So you start at some point on the circle, and whenyou reach the firstline somewhere inside the circle you have added another region by dividing an existing region into two parts, and whenyou reach the second lineyou have again added another region. And whenyou reach the other sideof the circle you have again added another region by dividing an existing region into two parts. So, when you draw the third line through the circle, there are three points at which one of the existing regions becomes two regions: when you reach each of the the first two lines you drew, and when you reach the opposite side of the circle. So we have: 3rd line drawn through circle ==> number of additional regions = 3

Now we generalize:

I think you can see where the analysis will go from here....When you draw the n-th line, you create n new regions-- (n-1) new regions, one for each time you intersect one of the previous (n-1) lines that you drew, and a last time when you reach the other side of the circle. So you can answer your specific question by making a simple table number of number of total number lines new regions of regions -------------------------------------- 0 -- 1 1 1 2 2 2 4 3 3 7 4 4 11 5 5 16 6 6 22

So rather than actually drawing all the way to 6 lines, we can quickly make a table, and get to that answer of 22. Or, we can turn this into a formula:

Then there are various mathematical techniques (which I won't go into here) for examining the sequence of the number of regions 1, 2, 4, 7, 11, 16, 22, ... to come up with the formula for the maximum number of regions formed by n lines: n^2+n+2 # regions with n lines = --------- 2

One way to do this is to see that the total number of regions is \(1 + (1 + 2 + 3 + … + n)\), and there is a formula for the sum from 1 to *n* (called the *n*th triangular number), namely $$ 1+2+3+…+n = \frac{n(n+1)}{2}$$ In fact, this is the same idea we used in counting handshakes. So our formula becomes $$r = 1 + (1 + 2 + 3 + … + n) = 1 + \frac{n(n+1)}{2} = \frac{n(n+1)+2}{2} = \frac{n^2+n+2}{2}$$ as before.

For one more derivation of the formula, we turn to this question from 2001:

Circle Regions What is the maximum number of regions you can have with n chords in a circle? I've already found: With 0 chords you have 1 field With 1 chord you have 2 fields; 1 more 2 4 2 more 3 7 3 4 11 4 etc. Now I need a formula and a proof why this is so. Can you help me?

Here the anonymous questioner has already made the table and seen the recursive description, and wants a formula. Doctor Anthony replied, first writing the recursion explicitly:

We can find a general formula for the number of regions when the interior of a circle is divided by n lines. Suppose the number of regions is given by f(n) when there are n lines drawn in the circle. Now draw one more line cutting all the other n lines. There are n points on the additional line, and so this line must traverse n+1 of the available f(n) regions, dividing each into two parts. It therefore adds n+1 more regions to those present. Thus f(n+1) = f(n) + n+1

As we’ve seen, each added line adds as many regions as the new number of lines. This could have been written as \(f(n)=f(n-1)+n\), but here we are thinking of obtaining the next number when we already have \(f(n)\). Now he writes this as a “telescoping” sequence of equations:

We can write f(n+1) - f(n) = n+1 and now form a succession of equations as follows: f(1) - f(0) = 1 f(2) - f(1) = 2 f(3) - f(2) = 3 ................... ................... f(n-1) - f(n-2) = n-1 f(n) - f(n-1) = n ------------------------- adding all the equations, f(n) - f(0) = SUM(1 to n) (note cancellation between lines) f(n) - 1 = n(n+1)/2 f(n) = n(n+1)/2 + 1

This is equivalent to what I said at the end of Doctor Greenie’s method.

Check if this is correct n=0 gives f(0) = 1 n=1 gives f(1) = 2 n=2 gives f(2) = 4 and these are correct. With n = 4 we get 4 x 5/2 + 1 = 11 regions. And so on.

Checking a few known answers is a good practice whenever you obtain a general formula!

Next week: What if we are given *n* points on the circle, and connect them with all possible chords?

In gathering information on how to count the diagonals of a polygon, I found this long discussion about a similar-sounding issue, which is hardly more difficult, yet far more complex. It was interesting to explore what the question means, and take it in different directions, on the way to the sort-of-simple answer.

Here’s the question, from 1999:

Diagonals in 3D Figures I am having great difficulty with math and require assistance. Could you help? I have the formula n(n-3)/2, but I don't know how to justify or prove it. Another thing is, could you please tell methe number of diagonals in various 3D shapes, such as a tetrahedron, cube and so on. Is there a formula for this or is it just coincidental?

I first explained the polygon formula, as we’ve seen it before:

Hi, Jamie. Let's thinkhow we can count the diagonals in a polygon. Pick any vertex; there are N ways to do that. Now pick any vertex to go to EXCEPT the two neighbors (and the point itself, of course). How many ways are there to do that? When you finish, you'll have counted every diagonal - except that you will have counted each one twice, once starting from each end. Taking that into account will give you the formula.

Can we extend this from two-dimensional polygons to three-dimensional polyhedra? I suggested an approach that leads to a final formula we’ll end up with:

Now, for a general polyhedron, the same method would work except for one detail:there can be any number of "neighbors" to any vertex. But you can use a similar method to find the TOTAL number of segments that can be drawn between any two vertices; then you can just subtract from your count the number of edges in the polyhedron. Since you can make different polyhedra with the same number of vertices but different numbers of edges,you can't give an answer based only on the number of vertices. Have you looked at atetrahedronand tried to count the diagonals? Try it - you'll find there aren't any. For acube, there will be two on each of the six faces, and three going through the center. See if that agrees with the formula you come up with.

Jamie replied with a few (incorrect) manually counted numbers, showing that he was not yet ready to derive a general formula:

Could you please help me? In a 3D object, the tetrahedron has 0 diagonals, the cube has 15 and then the pentagonal prism has either 28 or 32.Can you tell me the next 3 or 4 terms so that I can work out the sequence?Thank you very much.

As we saw last week, we can think of the number of diagonals of a polygon as a sequence, indexed by the number of sides, *n*: for *n* = 3, 4, 5, …, we have *d* = 0, 2, 5, …, from which you could guess the formula, or (far better) see a pattern and why it occurs. That is what Jamie was hoping for here. Unfortunately, an approach starting from specific numbers will not work, even if they are right.

I answered:

There are a couple of problems in trying to figure this out by making a list. For one thing,there really is no "sequence": you can't put all the possible polyhedra in an ordered list as you can with polygons, because there are lots of different ways to connect N vertices to make a shape. Secondly, even if you can see a pattern in the jumble of shapes you consider, it will be hard to be sure it's a real pattern if you haven't given thought to the reason for the pattern.It's much easier to find patterns if you look for a pattern in the way you count, rather than just make a list of numbers and then look for a pattern there.

We’ll see some examples below to demonstrate that we can’t just put polyhedra in a row with *n* = 3, 4, 5, … . But trying things out for small cases is still valuable:

Yet it will be worthwhile for you to make a list of shapes and try to count the edges and the diagonals of each, in order to get to know them. You'll want to get a feel forhow diagonals work, so it wouldn't be helpful for me to just give you a list -you need to do some counting on your own. But I'll start your list to give you some ideas on what to look for and how to count the diagonals.

I started with an example of each of several types of polyhedra, because each type has its own patterns.

Let's list several characteristics of each shape: the number of Vertices (V), Edges (E), Faces (F), and Diagonals (D). Tetrahedron V E F D --- --- --- --- 4 6 4 0 + / \\ / \ \ / \ \ / \ \ / __\__ + /____---- \ / +-------------+ Since every point is connected to every other point, there's nothing left to be a diagonal. Square pyramid V E F D --- --- --- --- 5 8 5 2 + / \\ / \ \ / \ \ / \ \ /---------\---+ / \ / +-------------+ The top point is connected to all the others, so the only diagonals are the two in the base. Triangular hexahedron (Twin tetrahedra) V E F D --- --- --- --- 5 9 6 1 + / \\ / \ \ / \ \ / \ \ / __\__ + /____---- \ / +-------------+ \ / \ / \ / \ / \ / \ / + The top and bottom points are connected to everything but each other; the "equatorial" points are connected to everything, leaving only the one diagonal. Notice thatthese last two polyhedra have the same number of vertices, but different numbers of edges. But the sum of the number of edges and the number of diagonals is the same for both, because that's just the total number of lines you can draw connecting any two points.

This is our first demonstration that V is not enough. I also hinted again towards our ultimate answer.

Continuing, with six vertices:

Octahedron V E F D --- --- --- --- 6 12 8 3 + / \\ / \ \ / \ \ / \ \ /---------\---+ / \ \ / +-------------+ \ \ // \ \ / \\ / + Each point is connected to every point except the opposite vertex; the three pairs of opposites form three diagonals. Pentagonal pyramid V E F D --- --- --- --- 6 10 6 5 + /|\ /|||\ / ||| \ / | | | \ / | | | \ / | | | \ /___|_-+-_|__ \ +- | | -+ \ | | / \| |/ +---------+ The top point is connected to all the others, so the only diagonals are the five in the base. Triangular prism V E F D --- --- --- --- 6 9 5 6 + / |\ / | \ +------------------+ | | | | | | | | | | | | | + | | / \ | | / \| +------------------+ The diagonals are all in the three sides. I've given youthree different polyhedra with six vertices. It gets a lot worse with more vertices!

Here are better pictures of the six examples, including the diagonals:

Again, I didn’t do this to generate numbers from which Jamie could deduce a formula, but (a) to demonstrate that polyhedra don’t form a sequence; (b) to show that the number of diagonals is not a function of V alone; and (c) to give some sense of what diagonals really are.

In particular, you might notice that the second, fifth, and sixth have only “**face diagonals**” lying on the “skin”, while the third and fourth have only “**body diagonals**“, diving through the middle of the “body”. Others, as we’ll see in a moment, have some of each.

I'm not sure why you gave two numbers for thepentagonal prism; the correct count, according to my formula, is 30: 5 in each base, 2 in each of 5 sides, and 2 reaching from each vertex on one base to vertices on the other base through the interior. The important thing to see is that you can't just look at V and tell me what D will be. You need to know at least two things about the shape.There is a formula that relates D+E to V, which is what I suggested to you in my last response; andanother that relates V+F to E. Putting these together, it turns out that D+F is determined by V. So if you know V and either E or F, you can figure out what D is.

Here are Jamie’s cube (square prism) and pentagonal prism:

I neglected to point out that Jamie missed one diagonal for the cube; there are 2 face diagonals for each of 6 faces, and 4 body diagonals (to the opposite vertex), for a total of 12 + 4 = 16.

We’ll get to the final formula I mentioned eventually, but I’ll mention here the two that we’ll be using. The first is simple: $$V = D+E$$ because every line from one vertex to another is either an edge or a diagonal, and it can’t be both.

The second was discussed in More on Faces, Edges, and Vertices: The Euler Polyhedral Formula: $$ V-E+F=2$$

Jamie wrote back:

Thank you very much for your help, it is greatly appreciated, I now understand more fully what my task actually is. Thank you once again for all your time with me.

But he hadn’t stopped thinking! Three days later, he wrote again:

Thank you for your faith in me, this is just a check, I have found your information onpyramidsuseful, and have come up with the formula (n-1)(n-4) d = ---------- 2 Is this correct? Just one other query; is there a link between the formula for pyramids and prisms, oris there one formula linking all 3D shapes together?

I answered:

Yes, you have the right formula for apyramid. I'm not sure how you found it, but there's a very easy way:all the diagonals will be in the base, since the apex is already connected to all other vertices; so you just have to put n-1 into the formula for diagonals of a polygon. You can do something similar for aprism: many of the diagonals will be in one or the other of the bases, and the rest can be easily counted because they go from a vertex on the top to a vertex on the bottom. I'm not sure there would be a direct link to the pyramid formula, but they certainly aren't unrelated either.

To be specific, the formula for a **pyramid**, using \(d=\frac{1}{2}n(n-3)\) for a polygon with *n* vertices, is (taking *v* as the total number of vertices) $$d=\frac{(v-1)((v-1)-3)}{2}=\frac{(v-1)(v-4)}{2}$$ as Jamie said. The formula for a **prism** (taking *v* as the total number of vertices, so that the number of vertices *in each base *is *b* = *v*/2) is $$d=2\cdot\frac{b(b-3)}{2}+b(b-1)=2b^2-4b=2b(b-2)=\frac{v(v-4)}{2}$$ These can be checked against my examples.

I had been hinting at the more general formula for some time, and did so again:

There is a very simple general formula for the total number of lines that can be drawn betweenany two of a set of n pointsin space, and this will be the sum of the number of edges of the polyhedron and the number of diagonals. You can easily derive your formula for a pyramid, or the one for a prism, from this by coming up with a formula for the number of edges. For example, for a pyramid, the number of edges is just 2(n-1), since there are n-1 edges in the base, and the same number of edges from the base to the apex.

We’ll get there!

Over the next few weeks, we discussed the derivation of the formula for prisms; Jamie had little knowledge of algebra yet, so there were a lot of details to go over, all of which I will omit here.

Then Jamie moved on to another specific case:

This is my last query; then I've figured the whole thing out. Can you tell me how many diagonals there are in adodecahedronand anicosahedron? Or can you tell me where on the Internet I can find that answer?

I answered:

It's time to introduce you to a different approach to the prism, which you can apply to these regular polyhedra.Regularitymeans that some or all features of a shape are true everywhere. A prism is not a regular polyhedron, because not all faces are the same shape and not all edges are the same length, and so on; but it is regular in one way:at every vertex you find the same number of edges, namely two going to adjacent vertices in the same base, and one going to the corresponding vertex in the other base. Luckily, that's the kind of regularity that helps most with this problem. You can use this to do the same thing you did originally in counting the diagonals of a polygon. If there are v vertices in all, you can pick any of them to start a diagonal, and then you go from there toany vertex except itself and the three other vertices to which it is connected, giving you (v-4) choices for each of the (v) starting vertices. Since you will have counted each vertex twice this way, the total number of diagonals is v(v-4) D = ------ 2 Sound familiar? You can use the same method for any regular polyhedron. See what you get! (You'll find that regular polyhedra are pretty boring.) Then see if you can actually visualize where these diagonals are and see that you counted them correctly.

In graph theory, the number of edges at a vertex is called its degree; when all vertices have the same degree, the graph is regular. So prisms and regular polyhedra have **regular graphs**.

Jamie applied these ideas to the two regular polyhedra he’d asked about (unfortunately getting some numbers wrong again):

Thank you for the e-mails. I find that I now understand it all, except for two small areas, which I hope you can help me with. The first is: using the prism formula and the notes you gave me, I worked out the number of diagonals. Can you check my data, see if they are correct, as I have two conflicting sources, the Internet and also an encyclopedia: Vertices Edges Faces Diagonals Dodecahedron 12 30 12 48 Icosahedron 20 30 20 160 If those are correct, which I believe them to be, I have come up with this logic, please tell me if I'm right. v(v-j) ------ 2 1. v This represents the total number of vertices in the polyhedron 2. v-j The total number of vertices minus j, where j is equal to the number of diagonals unable to be drawn to from any given starting vertex. For instance in the cube, from any given vertex, you are unable to draw diagonals to 3 vertices as they are connected with edges. Then you can't draw a diagonal to the vertex from where you started. So it would be the total number of vertices minus 4. This value would be (for all except two polyhedra) the shape of which the polyhedron is made from plus 1. The exceptions are the cube, where the 1 need not be added; and the octahedron, where it is needs to be added to 2. 3. 2 It is placed over two because by using this method you count each diagonal twice.

The **dodecahedron**, like the prism, has vertices of degree 3, so Jamie’s “*j*” is again 4; but there are 20, not 12, vertices, so we get $$d = \frac{v(v-j)}{2} = \frac{20(20-4)}{2} = 160$$ The **icosahedron** has 5 edges at each vertex, so “*j*” is 6; it has 12, not 20, vertices, so we get $$d = \frac{v(v-j)}{2} = \frac{12(12-6)}{2} = 36$$

I replied, neglecting to point out the errors in the numbers, and focusing on the thinking:

This sounds fine except for the details of how you find j. I would express the formula not in terms of j, which is not an obvious number, but in terms of something like thenumber of edges that meet at each vertex. It has nothing to do with the shape of the faces. You might like to make a chart for all the regular polyhedra and a few prisms, showing not only V, E, and F, but also the number ofedges per faceand the number ofedges per vertex. Then show what your "j" is, and see how it's related. Or just think carefully about what it means to be unable to draw a diagonal to a vertex.

Jamie’s ideas about calculating “*j*” appear to be based on the guess that it would be related to the shapes of the faces, and trying to find a relation between the numbers, rather than looking behind the numbers for a real reason. This is all too common when students have been taught to guess at patterns rather than derive formulas from processes, as we discussed here.

We never got to the end I was aiming for, because Jamie was too new to both algebra and geometry to be ready to see past the particulars. Let’s first finish what I suggested above, and then look more generally.

First, we can finish the formula for polygons in which each vertex has the same number of edges (**regular graphs**). If we call the degree of each vertex *g*, then Jamie’s “*j*” is just \(g+1\), and his formula becomes $$d = \frac{v(v-g-1)}{2}$$

Or, since the total number of “edge-ends” can be calculated as either \(2e\) or as \(vg\), we find that \(g = \frac{2e}{v}\), and the formula becomes $$d = \frac{v(v-\frac{2e}{v}-1)}{2} = \frac{v^2-v-2e}{2}$$ Checking this for our pentagonal prism with \(v=10\) and \(e=15\), we get $$d = \frac{10^2-10-2(15)}{2} = 30$$

If the graph of a polyhedron is **not regular**, then the best we can do is to use the idea I’ve mentioned since the start, that the diagonals consist of all **segments between vertices**, minus those that are **edges**: $$d = {v\choose 2}-e = \frac{v(v-1)}{2}- \frac{2e}{2} = \frac{v^2-v-2e}{2}$$

Amazingly, we get the same formula whether or not we assume the vertices are regular! This universal formula can also be written as $$d = \frac{v(v-1)}{2} – e$$

Since edges are often harder to count than **faces and vertices**, let’s use only the latter in a formula. Knowing that \(v-e+f=2\), we can replace *e* with \(v+f-2\). That yields this formula: $$d = \frac{v^2-v-2(v+f-2)}{2} = \frac{v^2-3v-2f+4}{2} = \frac{v(v-3)}{2}-f+2$$

From this, we can re-derive our previous formulas for prisms and pyramids:

There are 2 bases and *v*/2 sides, so that \(f = \frac{v}{2}+2 = \frac{v+4}{2}\); therefore, $$d = \frac{v(v-3)}{2}-\frac{v+4}{2}+2 = \frac{v^2-3v-v-4+4}{2} = \frac{v^2-4v}{2} = \frac{v(v-4)}{2}$$

There is 1 base with *v *– 1 sides, so that \(f = v\); therefore, $$d = \frac{v(v-3)}{2}-v+2 = \frac{v^2-3v-2v+4}{2} = \frac{v^2-5v+4}{2} = \frac{(v-1)(v-4)}{2}$$

It’s more natural to discuss these in terms of the number of sides in the base:

Since \(v = 2n\), we get $$d = \frac{v(v-4)}{2} = \frac{2n(2n-4)}{2} = 2n(n-2)$$

Since \(v = n+1\), we get $$d = \frac{(v-1)(v-4)}{2} = \frac{n(n+1-4)}{2} = n(n-3)$$

To check the formulas above, we can calculate these specific forms:

**Tetrahedron: **3-gonal pyramid, $$d = n(n-3) = (3)(3-3) = 0$$

**Cube: **4-gonal prism, $$d = 2n(n-2) = 2(4)(4-2) = 16$$

**Octahedron: ***v* = 6, *f* = 8, so $$d = \frac{6(6-3)}{2}-8+2 = 9-8+2 = 3$$

**Dodecahedron: ***v* = 20, *f* = 12, so $$d = \frac{20(20-3)}{2}-12+2 = 170-12+2 = 160$$

**Icosahedron: ***v* = 12, *f* = 20, so $$d = \frac{12(12-3)}{2}-20+2 = 54-20+2 = 36$$

These agree with our previous counts.

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