Conclusion to Half-Tau Month

Though pi day passed a few weeks ago, my brother made the interesting observation that this month is "Pi Month." It is March of 2014, or 3/14. Since I spent the month focused on trigonometry, I never had a chance to honor this joyous occasion until today.

Interestingly, pi does play a huge role in trigonometry. From wrapping functions to sine curves to angle measurements, pi is always popping up. Though this is kind of interesting, trigonometry is also one of the areas where tau really shines. Having just finished topics such as trigonometry, polar coordinates, and wrapping functions, I have found that it is a real struggle to use pi. I found myself converting most of my problems to tau before solving them just because pi made it too confusion.

No video describes these sorts of issues better than Vi Hart's "Pi is (still) Wrong" video, which gets into some of the issues involved with using pi, one of these being trigonometry.

I would also like to make you all aware that next month is Math Awareness Month. The theme this year is Mathematics, Magic, and Mystery, in part to honor the centennial of Martin Gardner's birth. At www.mathaware.org, there is a poster with 30 squares on it to represent the 30 days of April. On each day of the month, the next square becomes active. I will try to keep an eye on these webpages, as I will probably use April to comment on the topics posted there. Also, April 1st is a page on magic squares, and I am extremely honored to be hosting that day.

Rediscovering Trigonometry Part 1: Addition and Subtraction Formulas

In January, I did a series with some heavier mathematics, involving lots of number theory, combinatorics, and some fairly difficult manipulation. Last month, we took a little rest from this by talking a bit about casino games and optimal strategies. The mathematics we used was more probability and game theory. This month, I thought we could explore something completely different that is also extremely interesting. It is taught in school (much of the material in this post actually comes straight from my math class), but maybe not in a fun, thought-provoking way. This month, we will dive into trigonometry.

If you want to brush up on your basic foundational trigonometry, please click here. It is my post on the Law of Cosines, but also explains at the beginning what the sine, cosine, and tangent functions are. I will make sure to prove any other trigonometric identities we use in this series (or link to a proof), so all you need to know is what a sine, cosine, and tangent is.

Now we are ready to derive our first identity. We will try to figure out a simpler way of writing sin(α+β). There are many proofs of this, but I am using one that I found at www.themathpage.com. Let's start with a diagram to help explain this idea.

Create the line AB.

Rotate it to point C, creating the angle α.

Rotate it to point D, creating the angle β.

Draw DE perpendicular to AB.

Draw DF perpendicular to AC.

Draw FG perpendicular to AB.

Draw FH perpendicular to ED.

We are trying to figure out the sine of α+β, which can be written as (ED/DA). First, let's determine the value of the angle HDF (which has already been written in for us).

Note that the lines HF and AG are parallel, with a transversal (a line that intersects both lines) of AF. In geometry, there is a theorem called the Alternate Interior Angle Theorem, that essentially tells us that in this circumstance, the measure of angle GAF is equal to the measure of angle AFH. In other words, angle AFH has a measure of α. Noting that angle AFD has a measure of 90 and AFH has a measure of α, that means angle HFD would have a measure of 90–α. Since a triangle has 180 degrees, having two angle measures in a triangle is enough to get the third. Triangle HFD has an angle of 90, an angle of 90–α, and one more angle that creates a sum of 180 degrees. Do the math and you will find that the angle must be α, and thus, angle HDF has a measure of α as depicted in the diagram.

Now we can try to find the formula. Let's start with the following equality:

ED = GF + HD

Note that we are trying to find what (ED/DA) equals. Because of this, let's divide both sides by DA.

(ED/DA) = (GF/DA) + (HD/DA)

Let's complicate some terms on the right hand side. Multiply the fraction on the left by (AF/AF) and the one on the right by (FD/FD)

(ED/DA) = (GF/AF)(AF/DA) + (HD/FD)(FD/DA)

(ED/DA) is simply the sine of α+β. Looking back at the diagram, (GF/AF) would be the sine of α. (AF/DA) would be the cosine of β. (HD/FD) would be the cosine of α. (FD/DA) would be the sine of β. This gives us the following formula:

sin(α + β) = sinα cosβ + cosα sinβ

This is a beautiful result. The sine of the sum of two quantities can just be pulled apart so easily. This is really cool on its own, and furthermore, it ends up being the foundation of tons more identities. Let's say we wanted to find the sine of α–β. We could just plug (-β) in for β and see what we get.

sin(α – β) = sinα cos(-β) + cosα sin(-β)

These negative terms can be analyzed in a number of ways, but the easiest way to do it is to look at the graphs of the sine and cosine functions.

Graph of y = sin(x)

Graph of y = cos(x)

Notice how the sine graph is symmetric over the origin. Any sin(x) will equal -sin(-x) and any sin(-x) will equal -sin(x). On the other hand, the cosine graph is symmetric over the y-axis. Any cos(x) will equal cos(-x) and -cos(x) will equal -cos(-x). This enables us to simplify the formula much more.

sin(α – β) = sinα cos(-β) + cosα sin(-β)

sin(α – β) = sinα cosβ + cosα (-sinβ)

sin(α – β) = sinα cosβ – cosα sinβ

Fantastic. What if we wanted to find the cosine of the sum of two angles? Going back to our original diagram, we can start with the following expression:

EA = GA - FH

Dividing through by AD creates the desired left hand side.

(EA/AD) = (GA/AD) - (FH/AD)

Multiply the term on the left of the right hand side by (AF/AF) and the right term by (DF/DF) to get:

(EA/AD) = (GA/AF)(AF/AD) - (FH/DF)(DF/AD)

And turning this into the appropriate functions gives:

cos(α + β) = cosα cosβ – sinα sinβ

Doing the same thing as before gives a subtraction formula of:

cos(α – β) = cosα cosβ + sinα sinβ

Finally, let's try to create a tangent formula. Knowing that dividing the sine by the cosine will give the tangent, let's try dividing the two formulas by each other.

sinα cosβ + cosα sinβ

cosα cosβ – sinα sinβ

Divide each term through by cos cosβ.

(sinα cosβ)/(cosα cosβ) + (cosα sinβ)/(cosα cosβ)

(cosα cosβ)/(cosα cosβ) – (sinα sinβ)/(cosα cosβ)

Many terms will now cancel.

(sinα)/(cosα) + (sinβ)/(cosβ)

1 – ((sinα)/(cosα))((sinβ)/(cosβ))

Turning the sine and cosine ratios into tangents gives a formula of:

tanα + tanβ

1 – tanα tanβ

And there's the formula! If you wanted to create a subtraction formula, you would just note that the tangent function is symmetric over the origin (or it is an odd function), and plug in the necessary values to get:

tanα – tanβ

1 + tanα tanβ

I think that these are really cool formulas. But they are also very practical. For instance, let's say you want to find the exact sine of 75°. We know that sin(45) = (√2)/2, sin(30) = 1/2, cos(45) = (√2)/2, and cos(30) = (√3)/2. So, this can be determined with the sine angle addition formula.

sin(75) = sin(30 + 45) = sin30cos45 + cos30sin45 = (1/2)((√2)/2) + ((√3)/2)((√2)/2) = (√(2) + √(6))/4

Knowing that the Greeks did some manipulation with the regular pentagon to find an exact sine of 72°, we can plug all of this information into the sine angle subtraction formula to find the exact sine of 3°. We can then eventually make our way down to 1°, which gives us the ability to generate the sines of all of the angles on the trig tables that mathematicians now use today.

Next week, we will use these formulas to create more generalizations that will make many of these trig table computations much much simpler. And give us some really pretty formulas in the process.

Happy End Problem

As this is the last post of the year, I thought it would be appropriate to end with a post on the "Happy End Problem." I will explain the problem, which is pretty cool in itself, and then talk a little bit about the history behind it.

The initial question was if any five points were placed on a plane with no three of them in a straight line, will four of those points always form a convex quadrilateral? For example, in the following image:

The following convex quadrilateral can be formed:

Will this always work? As usual, I encourage you to grab a scrap piece of paper and try out a few examples. Have fun with it. Get creative! You will end up finding that no matter how you position the five points, you cannot get a combination without a convex quadrilateral.

Why is this true? In fact, there is a very easy way to prove it. Let's analyze three cases.

The first case is the top left one in the red, where the five points form a convex pentagon. In this instance, connecting any four of the points will form a convex quadrilateral by nature.

The second case is the top right one in the blue, where one point is located in between the four outside points. The illustration shows the inside point being included in the quadrilateral, but it could have just as easily been made as just the four outside points. This will continue to work for any combination of this nature by logic.

The third case is the bottom one in the yellow, where two points are enclosed in a triangle. When you draw a line between the two center points, two of the outside points will end up on one side and one will be on the other. Using the two outside points as your third and fourth vertices will form a quadrilateral without flaw.

Now, you might be wondering if one can prove a similar case with a convex pentagon. Could it be done with six? Seven? Eight? Turns out, nine points are required for it to work every time. As you can see below, eight points is just one too few.

What about convex hexagons? Or heptagons? Or octagons? Or chiliagons (1000-sided polygons)? Well, what many mathematicians will do from here is look for a formula to figure out how many points are required for a given n-gon. We know that for a triangle (n = 3), just 3 points are needed (all triangles are convex). For a quadrilateral (n = 4), we proved that 5 points are needed. For a pentagon (n = 5), I mentioned that 9 points are needed. Do you see the pattern?

3, 5, 9, ...

It is not easy to spot at first, but what if I subtract one from each of those terms:

2, 4, 8, ...

They are all now powers of two! This pattern seems to fit the formula A_n = 2^n-2 + 1. Plugging six in for n would give:

A₆ = 2^6-2 + 1
A₆ = 16 + 1
A₆ = 17

This formula predicts that seventeen points would be required for a hexagon. Mathematicians would then work to try to prove that this is the case. Further, they would try to prove that for any value of n, the A_n formula holds true.

George Szekeres (1911-2005; a Hungarian-Austrailian mathematician and analytical chemist) and Esther Klein (1910-2005, another Hungarian-Austrailian mathematician) worked together to prove that all values of n will have a finite A_n output (there will be a number of points that creates the ability for a convex n-gon to be formed), but they could not get this bound down to the formula above. Soon after this proof was published, Szekeres and Klein married each other, which inspired the name "Happy End Problem."

Paul Erdös (1913-1996; a Hungarian mathematician), possibly one of the most influential of the twentieth century), was able to prove successfully that with 71 points, a hexagon can always be drawn.

Sixty years later, Ronald Graham (born 1935; a Californian mathematician) and his wife, Fan Chung, decided to take a swing at the problem. While on a plane ride to a math conference in New Zealand, they were able to lower Erdös's bound to 70 points, which doesn't sound like much, but it brought the problem back into the minds of mathematicians. It was also ironic that another achievement pertaining to the Happy End Problem was from a couple.

Daniel Kleitman (born 1934; an applied mathematician at MIT) and Lior Pachter (born 1972; an Israeli mathematician and molecular biologist at Berkeley College) worked together to lower the upper bound to 65 points. The number was then lowered to 37 points, and has yet to be lowered further.

Although lots of progress has been made on this problem, the overarching proof still has not yet been found. There has been no counterexample to the A_n formula, and there has certainly been no guaranteed formula to generate the future values. People often wonder what a mathematician actually does for his/her job. A big part of it is trying to figure out the answers to these unsolved problems, which can often be understood by the average person. Try playing around with it and you might make a discovery too.

Isosceles Triangle Theorem

The field of geometry is comprised of many different types of questions. Some are construction based, such as “what is the area of this circle?” On the other hand, many are proof based, like “why are these two triangles congruent?” This is slightly different from the proofs I normally discuss; proofs I normally post are very generalized theorems while these questions are more similar to a specific algebra or arithmetic problem.

When writing a geometric proof, many different theorems come into play. One must use the generalized theorems, properties, and postulates to arrive at a conclusion. Let’s try a simple proof just to demonstrate the nature of these theorems. Let’s prove that triangle ABE is congruent to triangle CDE.

First, we would say that it is given that segment AC is parallel to segment BD and that segment AB is parallel to segment CD. We would then say that ABCD is a parallelogram by the definition of a parallelogram (which requires two sets of parallel sides). The definition of a parallelogram also requires AB and CD to be congruent. The vertical angle theorem can be used to say that angle AEB is congruent to angle CED. The alternate interior angle theorem says that angle ABE is congruent to angle DCE. The angle-angle-side triangle congruence postulate then concludes that triangle ABE is congruent to triangle CDE.

As you can see, there are many steps in this proof, and each one uses a different definition, theorem, or postulate. In high school geometry classes, students are told these theorems and postulates, and expected to memorize them for future examples. This makes geometry boring and pointless, when it can be quite fascinating. A way to easily spice up geometric proofs is to actually prove the theorems before they are used in class. If Euclid could do it, then we can do it.

A fun one to prove is the Isosceles Triangle Theorem. This theorem states that when a triangle has two congruent sides, it also has two congruent angles. This can be proven in a similar way as the congruent triangle question I posed earlier. Take an isosceles triangle:

If you were to bisect that top angle, it would create two new triangles. Since the original triangle is isosceles, it is given that the top left segment is congruent to the top right segment. The definition of a bisection (cutting an angle in half) states that the left part of the top angle is congruent to the right part of the top angle. The reflexive property of congruence states that the middle segment is congruent to itself. By the side-angle-side triangle congruence postulate, the left triangle is congruent to the right triangle. And finally, by CPCTC (common parts of congruent triangles are congruent), the bottom left angle is congruent to the bottom right angle.

This sort of geometric proof language sounds extremely long and boring. However, finding uses for it such as proving the Isosceles Triangle Theorem can make it a little more fun. Among many things, I think that schools should teach the reasons behind these theorems to make it more logical and fun to apply them to class.

Math in the News: Giving Students More Independence

About a month ago, I was speaking at a TEDx conference that was themed around education. During these TEDx conferences, there are always live speakers as well as videos chosen by the organizers that fit well with the occasion. One of the videos we watched really got us thinking. Here it is:


This talk is not directly mathematical. It isn't meant to be a math education talk. But, these points apply to math education as well.
For example, the mathematician Euclid had nothing. He pretty much was starting from scratch, just like the kids described in the talk. So what did he do? Well, after creating five axioms (foundational ideas that can be concluded with basic logic), he started asking questions, and answering them with mathematical proofs. He would then think about other questions, and find ways to answer them using everything he had discovered. This was the basis of his book series called Elements.
Students should be able to approach math in a similar way. A teacher could lay out five axioms (or develop them with the students), and then back off. He or she might also provide terminology (line, triangle, square, circle, angle, bisect, trisect, etc.), but the students can discover the rest. By learning math this way, they will understand everything they are doing so much better because they created it. For more on that topic, check out my Capstone Research Paper (link is on the top of the page), and scroll to "Chronological Cognition."

Law of Sines

A few months ago, I posted the proof of the Law of Cosines, which is an extremely important aspect of trigonometry. If you do not know what sines and cosines are, it is a very easy concept. Click here to view the post discussing that (which is also the post with the proof of the Law of Cosines).

But this law only works if you are given all three sides of the triangle or two sides and the enclosed angle. What if you are given two sides and a non-enclosed angle, or two angles and one side (three angles isn't enough information to generate side lengths)? How could you approach this problem?

This is where you use the Law of Sines. This law goes as follows:

This was taught to me last year in school, and I immediately wondered what the proof was. Though the law of cosines one was a bit clunky, I found that this proof was quite simple and elegant. So, I thought that it would be great to share.

Since it would require many diagrams, I thought it would be easier to just watch a video of it. It is pretty short, and explains the proof well.

A big part of the reason why most of the cool stuff I post isn't taught in school is that it is not mandated in the curriculum or test standards. Of course, I do believe there are changes that need to be made to these (click here for my Capstone research paper explaining those). However, the Law of Sines is something already taught in school. Same with the Law of Cosines.

These proofs, especially the sine one, fit right into the curriculum. The Law of Sines is already being taught, so why not take an extra 5 minutes to explain the proof? Or even better, explain the basic thought process behind the proof and have the students generate the formula (which works really well for the Quadratic Formula as well). This increases the students' ability to understand and apply the concept, as well as making it fun and interesting. On top of that, the common core standards do want students able to "construct viable arguments," which is the whole purpose of proofs. I think that this proof is not only interesting, but shows that cool math stuff can be integrated into the classroom while keeping it relevant and obedient to the standards.

Don't Borrow Tens Ever Again in a Subtraction Problem!

When the traditional method of subtraction is taught in school, it can often be quite confusing, especially when the number you are subtracting from has lots of zeros. It is a pain to go borrow the tens from other numbers, especially when they are ones or zeros. But, there is an easier way to approach this that isn't taught in class.

In geometry, the word "complimentary" is used to describe the relationship between two angles that sum to 90°. For instance, 36 and 54 are complimentary angles. Similarly, the compliment of 36 would be 54 when talking about geometry.

A geometric application of complimentary angles (and a funny picture)

With arithmetic on the other hand, a number's compliment is its difference from the power of ten above it. So, the compliment of 36 would be 64 (100 - 36 = 64), or the compliment of 473 would be 527 (1000 - 473 = 527). This is more practical for solving arithmetic problems than the distance from 90.

One-hundred or one-thousand are very difficult numbers to subtract from. These are ones where you need to keep turning zeros into nines until you reach that first digit, and then figure out the problem from there. This can be quite the pain. But, there is a shortcut for these examples.

All you have to do is subtract the last digit of the number from ten. This is the last digit of the compliment. Then, subtract all of the rest of the digits from nine, and place them in their corresponding place value. For instance, let's find the compliment of 36.

10 - 6 = 4
9 - 3 = 6

100 - 36 = 64

Pretty easy, right? Let's try it with 473.

10 - 3 = 7
9 - 7 = 2
9 - 4 = 5

1000 - 473 = 527

This can easily be extended to numbers in the thousands, millions, billions, and more as long as you can keep track of the digits. For example, 5647823 would have a compliment of:

10 - 3 = 7
9 - 2 = 7
9 - 8 = 1
9 - 7 = 2
9 - 4 = 5
9 - 6 = 3
9 - 5 = 4

10000000 - 5647823 = 4352177

Couldn't be easier! This method is basically just remembering the fact that no matter what the situation, you will have borrowed a ten from each zero, making them all nines except for the very last one which will remain a ten. Try a few subtraction problems yourself and you will see why it works.

This can be applied to geometry in a way as well. To subtract quickly from ninety, you still subtract the last digit from ten, and the first one from eight (one less than the nine). So, the compliment of 29° is:

10 - 9 = 1
8 - 2 = 6

90° - 29° = 61°

For supplements of angles (the angle's difference from 180), you can do a similar technique as well. Subtract the last digit from ten and the first from seventeen. For instance, the supplement of 48° is:

10 - 8 = 2
17 - 4 = 13

180° - 48° = 132°

In fact, any number ending in zero(s) can be subtracted from just by altering this method. I have found this extremely helpful when performing mental math (three-digit and four-digit squaring requires you to quickly identify how far you are from the nearest hundred/thousand, which often needs compliments). It is also very practical. When you give the cashier a hundred dollar bill, they are usually impressed when you tell them the change before they have time to punch the bill into the register. I'd recommend practicing this technique because it is useful and also quite a bit of fun.

Math in the News: Playing with Pentagonal Tilings

This last weekend, I attended MathFest 2013, which is held by the Mathematical Association of America in Hartford, Connecticut. I gave a talk on magic squares, and got to see many other fascinating speakers. Many of my future posts will be content I learned at this convention.

One of the speakers was Frank Morgan, who talked about tiling the plane with different polygons, primarily pentagons. During the presentation, he mentioned that he wrote a news article about it in the Huffington Post. Since the presentation was so interesting, I thought this article would be a perfect "Math in the News" post.

http://www.huffingtonpost.com/frank-morgan/bees-honeycomb-mathematics_b_1318258.html

The Law of Cosines

At some point during your high school math experience, whether it be during an Algebra II, Geometry, Trigonometry, or Precalculus class, you probably were presented with the Law of Sines and the Law of Cosines. You found that the sine and cosine ratio could only solve right triangles, and that you needed something more to solve non-right triangles. So, two formulas were slapped onto your textbook, and you had to apply them.

Today, I am just going to focus on the Law of Cosines, and save the Law of Sines for a future post. This formula is one where you probably wondered why it worked. You might be asked to prove something using this formula, but how can you comfortably do that without being sure of the formula in the first place?

First, I need to lay a little foundation. Take a right triangle:

First off, we should know the Pythagorean Theorem. Click here for an explanation and proof of it.

Let's say we are dealing with the bottom left angle. The side opposite to this angle has a measure of 4, and will be referred to as the "opposite" side. The longest side has a measure of 5, and will be referred to as the "hypotenuse." The remaining side has a measure of 3, and will be referred to as the "adjacent" side, since it is adjacent to the angle.

Now, let's make sure we are on the same page with terminology. The sine of this angle is the opposite side over the hypotenuse. 4/5 is 0.8, so the sine of that angle is 0.8. The cosine of this angle is the adjacent side over the hypotenuse. 3/5 is 0.6, so the cosine of that angle is 0.6. Tangents will not be needed in this post, but it is the opposite over the adjacent. These ratios are commonly remembered by the SOH CAH TOA acronym.

There is a pretty cool identity found in the sine and cosine ratios. Take the two ratios that we just found for the 3-4-5 triangle:

sin(x) = 0.8

cos(x) = 0.6

Now, plug those into the following expression:

sin²(x) + cos²(x)
0.8² + 0.6²
0.64 + 0.36
1

Interestingly, this sum always turns out to be one. By using those ratios and the Pythagorean Theorem, you can prove that for all angles, the square of the sine plus the square of the cosine is one.

With this information, we can prove the Law of Cosines. The formula slightly resembles the Pythagorean Theorem, so we will try to keep that in mind when proving it.

Each of the derived measurements are just rewritten forms of the sine and cosine ratios. For example, the cosine of C is that adjacent side over b, so multiplying both sides by b yields the measure of the adjacent side: b • cosC.

Let's look at the right triangle with hypotenuse c, and solve the Pythagorean Theorem. Since the Law of Cosines is similar to the Pythagorean Theorem, this might give us a start.

[a - (b • cosC)]² + (b • sinC)² = c²
a² - 2abcosC + b²cos²C + b²sin²C = c²
a² + b²cos²C + b²sin²C - 2abcosC = c²
a² + b²(cos²C + sin²C) - 2abcosC = c²

Wait a minute - what is inside the parentheses? We have cos²C + sin²C. But we proved earlier that that is always equal to one. So, we can substitute one in for that sum, and see where that takes us.

a² + b²(cos²C + sin²C) - 2abcosC = c²
a² + b²(1) - 2abcosC = c²
a² + b² - 2abcosC = c²

And we end up with the Law of Cosines: a² + b² - 2abcosC = c². I think that this proof is a pretty cool one, considering that it brings so many other neat identities into play.

History of Math: Irrational Numbers

I normally end each month with a post on a mathematical story from history. Since I've spent the last month discussing the Pythagorean Theorem, I thought I'd finish it off with a story from the life of Pythagoras.

Pythagoras was born in 570 BC in Samos (which is now in Vathy, Greece). Throughout his life, he dabbled in philosophy, mathematics, music, and religion (he actually was the founder of the religion known as "Pythagoreanism"). He founded an organization (sort of like a school) where lots of ideas were developed. It is unclear if many mathematical and philosophical ideas of the time were directly from him or from one of his students.

The Pythagorean Theorem was used by Babylonian and Indian mathematicians who preceded him. However, he or one of his students was the first to determine a proof of it. Prior to that, the theorem would just be a conjecture (for an idea to be considered a theorem, it must be proven; otherwise, it is a conjecture).

Pythagoras had a belief that all numbers could be expressed as a fraction or terminating/repeating decimal. This held true until Hippasus, one of his students, tried solving the Pythagorean Theorem with legs 1 and 1. This yielded a hypotenuse of length √2.

1² + 1² = c²
1 + 1 = c²
2 = c²
√2 = c

Hippasus then tried to write this number as a fraction. During his attempts to do this, he ended up proving that the number can not be written in this way (this might not be Hippasus's exact proof, but click here to see a proof of this).

After Pythagoras found out, he was outraged. He did not want to accept the fact that irrational numbers exist. He tried proving Hippasus wrong, but could not do it. Because of this, Hippasus got to take a "boat ride" in the lake, where he was thrown overboard and drowned.

I found it interesting that in that day in age, people reacted this way to the discovery of irrational numbers. Though nobody got killed when complex numbers were found in 1539, there was lots of discomfort. Newton and Euler are both recorded as calling them "impossible" numbers, and the letter i stands for imaginary, since this number could not possibly exist. Gauss was the first to recognize complex numbers as a real thing.

Fun Puzzle with the Pythagorean Theorem

Normally, I spend the third week of the month discussing a practical application of a cool mathematical concept. However, my post from last week fit that standard perfectly. So, I chose to give you a practical use of the Pythagorean Theorem, but less on a applicable level and more on a personal level. This is an identity that can be turned into a puzzle for your friends. It can also be used as a proof of the Pythagorean Theorem. Enjoy!

Math in the News: The Pythagorean Theorem in Football

I mentioned last week that I wanted to focus this month on the famous Pythagorean Theorem. Since I usually reserve the second week for a "Math in the News" segment, I was looking for an example of the Pythagorean Theorem in the news.

Though I couldn't find a controversial article about it, I did find a fantastic video that showed its applications to football. And it was done by the National Science Foundation in their news section, so I think it counts. Click here to see the video.

My friends were just joking with me a few weeks ago about how in a sports game, players wouldn't actually be thinking about the math behind what they are doing. It's true that they probably aren't solving for the hypotenuse of a triangle in this dynamic moment. However, they actually do need to take an educated guess as to what angle and speed they should be running at. This does require number sense, which is definitely mathematics. The actual math might not be applied directly, but it is definitely present.

When I talked about connecting game theory with penalty kicks in soccer, this example actually uses the mathematics. Teams do hire statisticians to analyze the skills of the players on both sides, and will give their players advice on what to do. I'm sure coaches of football and other sports do this as well. Why else would they watch the opposing team's previous games to prepare themselves?

I had never thought the Pythagorean Theorem could be applied that way, and it is another reminder of the influence that mathematics has on our lives.

Proof of the Pythagorean Theorem

This month, I decided I would focus on one of the most famous theorems in all of mathematics: The Pythagorean Theorem. Though it sounds kind of dull on the face, it has a lot of extremely interesting things within it.

First, let me go over what it is. If you take a right triangle, and label the measure of its shortest side a, its middle side b, and its longest side c, then you can count on the fact that:

a² + b² = c² 

This can be used in geometry problems, like to find the longest side, or the hypotenuse, of a triangle with shorter sides, or legs, 3 and 4. You would simply plug 3 in for a, 4 in for b, and solve for c.

(3)² + (4)² = c² 

9 + 16 = c² 

25 = c² 

5 = c

The first thing I wondered when seeing this theorem is why it is true. It seems odd that any right triangle’s measurements must fit this criteria. And I never found a proof I liked until extremely recently. So, I would like to share it.

Take the following diagram:

We are going to try to figure out the area of this square. The simplest way to do it would probably be to square the square’s side. This length would be a + b.

(a + b)²

a² + 2ab + b²

Another way we could get this result is to find the area of the inside square, and then the area of the four triangles surrounding it. The inside square has length c, and the outer triangles have a base of b and height of a. So, we would get:

c² + 4(ab/2)

c² + 2ab

Since these two quantities are both the area of the same square, we can set them equal to each other. After simplification, we would get:

a² + 2ab + b² = c² + 2ab

a² + b² = c²

And we end up with our Pythagorean Theorem. I was so intrigued to see that this proof was so quick and simple, as well as pretty interesting.

Eratosthenes: The First Estimate of the Earth's Circumference

You might have thought that the Earth was proven round by Christopher Columbus in 1492. However, it was actually proven almost 2000 years earlier by a mathematician named Eratosthenes.

Eratosthenes was born in 276 BC in Cyrene, which was then a Greek colony. It would now be considered part of northern Libya. He was educated at Plato's school in Athens and went on to become chief librarian at the University of Alexandria.

Aside from his other contributions, he came up with a very close estimate of the circumference of the earth. First, he found out that at noon on the summer solstice, people in the Egyptian city of Syene noticed that there were no shadows from the sun. This is because Syene is very close to the Tropic of Cancer. This was confirmed by some tests he did by looking for brightness contrast inside of deep wells.

In his hometown of Alexandria, thought to be on the same meridian as Syene, he measured the angle of the sun's shadow on a pillar at noon of the next summer solstice. He found a 7° angle. This meant that when a circle is drawn to represent the circumference of the earth, Alexandria would be 7° around the circle.

Syene was known to be about 500 miles from Alexandria, and 7° is about one fiftieth of a circle. So, multiplying 500 x 50 gave Eratosthenes his estimate of 25,000 miles. This is a very impressive estimate, considering the modern distance is 24,901 miles.

It is extremely cool that you can determine the circumference of such a big thing without much technology at all. Just with a few mathematical principles, Eratosthenes could solve this daunting problem.

Math in the News: Facebook Math Problem

A few weeks ago, I suddenly opened up my Facebook page to see that I was tagged by several people in a post about this article. Once I saw it, I immediately wanted to discuss it on Cool Math Stuff.

First, try to solve this math problem:

6 ÷ 2(1 + 2) = 

Even though it seems simple, there is lots of debate about the answer on social networking sites. Click here to view the article.

When I first saw the problem, I thought the answer was undoubtably 1. When I did the problem, I had multiplied by two while I was simplifying the parentheses. So, I assumed that people who got 9 as the answer were forgetting the order of operations.

Before I talk about the article, I was excited to see that this was a debate between people. Any time where the general public is talking about mathematics, especially in a argumentative way, is a huge accomplishment for society.

Reading the article, I was intrigued to find myself in the minority. I did not think of the parentheses as a substitute for the multiplication symbol, but a quick way to write two times the quantity one plus two.

What if the problem was written like this?

6 ÷ 2 • (1 + 2) =

If the problem were written like that, I would have gotten 9 as my answer. Since the 2 was separated from the parentheses, "Please Excuse My Dear Aunt Sally" would put the division first and then the multiplication.

However, without the dot there (more formally known as an interpunct), I do feel that the answer should be one. It seems like the most practical perspective on the problem.

Take an expression like 2a. It is implied that the 2 and the a are a single quantity. If you were using it in an actual problem, it is far more likely that you would be dividing, say 8, by the whole quantity 2a than 8 by 2, and then multiplying this by a.

Similarly, let's say you had 2(a + 3). Again, there would be no doubt that you should distribute the 2. This problem would then become 2a + 6, which seems like the correct thing.

If you ended up with 1 ÷ 2(a + 3), there would be no practical case where the one were to be divided by the two first.

Also, I think that the order of operations was put into place to make algebra and geometry more structured. For instance, say you had rhombus where the diagonals cross, or bisect, at a (3x³ + 9)° angle, and you had to find the measure of the sides which were of length 25 - 2x². As you probably know, a rhombus is a quadrilateral with all four sides equal (a square has all four sides and all four angles equal).

There is a theorem that states that the intersection of the diagonals of a rhombus is always equal to 90°. I think this is pretty cool on its own, to know this angle all the time. So, we can assume that 3x³ + 9 must equal 90°.

3x³ + 9 = 90
3x³ = 81
x³ = 27
x = 3

We must then plug this into the expression equal to the side measure to find our answer. First, let's look at the expression.

25 - 2x²

What are we being asked to do? Since two is a coefficient for x², we would square the x first, then multiply by 2, and then subtract that from 25. This would be agreed upon by all algebra teachers. What happens if we plug the 3 back in?

25 - 2(3)²

If there was no order of operations, we would probably do this problem from left to right. This would give us:


25 - 2(3)²

23(3)²

69²

4761

You can probably already tell that 4761 is way bigger than this side length would be intended to be. Even if you didn't make this estimate, this attempt at the simplification was nowhere close to the agreed order. By implementing the order of operations, there is no debate that the order should be PEMDAS. This clarifies lots of algebra when you are substituting terms into equations.

What if the expression were 108 ÷ 2x²? I still think most of us would agree that the 2x² is meant to be its own term, and would therefore be divided as a quantity into the 108.

But after substitution, we get the following:

108 ÷ 2(3)²

108 ÷ 2(9)

If we did the 108 ÷ 2 first, we would end up with the wrong answer. Even though the surface definition of PEMDAS would ask for this to be done first, it is not a practical approach to the problem.

However, I did see that the SAT or ACT would expect students to receive 9 as an answer. Since there is clearly not a correct perspective to have, I would encourage you to comment what you thought the answer to be and why. Just like with Pi vs Tau, this is a post that is actually a lot of fun to debate about.

The Museum of Mathematics

I have been trying to find a news story once a month that pertained to mathematics that I could discuss. Last month, I ended up using a website I found, but this time, I actually got a news story. On Sunday, I was told by multiple people that the Museum of Mathematics was on CBS, which was exciting for me. Click here to see the news story.

I have met Glenn Whitney on a few occasions, including at Gathering 4 Gardner in Atlanta (click here to see a post I did about that experience) and the World Science Festival Street Fair in New York City. Each time I see him there, they have set up their Math Midway, which features many fascinating ways to have fun with mathematics, including the tricycle with square wheels and a machine that solves quadratic equations by the crank of a lever.

I visited MoMath on December 27th, which might have been a mistake on my part because it was packed with people. However, I was happy at the same time that a whole room of people were going to walk out seeing mathematics in a new light.

On my blog, my posts are mainly about pure mathematics. Pure mathematics analyzes math simply for the sake of analyzing it, while applied mathematics looks for its applications to science, engineering, finance, and society. I do have many posts labeled Practical, but even these are strictly mathematical. I never introduce science and engineering concepts (mainly because I don't know them).

MoMath gave me the opportunity to learn how math can be applied to other fields. I have ridden the square wheeled tricycle, which is a way to apply mathematics to transportation, in a fun and creative way. Using some geometry, you can find a path for any shaped wheels, which I found pretty cool.

If you are interested in pure mathematics, that is what this blog and many others cater to. If you are interested in applied mathematics, go to MoMath. It really is a fun way to look at math that you never would have done before.

Why Does πr^2 Work?

Recently, I have been alluding to the pi vs tau argument, which is debating the proposal of replacing pi with tau (the equivalent of 2π). Since both sides have pretty convincing points, it is a fun thing to talk about.

Pi fans argue that by changing pi to tau, we would ruin the formula for the area of the circle. It would go from:

A = πr^2  –>   A = 1/2τr^2

As you can see, the formula looks a lot sloppier.

Tau-ists rebut this by saying that the proof of this formula requires one to multiply 1/2 by 2πr^2, thus proving the significance of 2π. I mention this a lot, but I have never actually proven it.

Rather than writing out the proof, I think it would make more sense to watch this YouTube video. It made it a lot more interesting for me.

I find this proof fascinating on its own. Also, you may have noticed when the 2π was present and was cancelled out by the 1/2. For the pi vs tau argument, this specific instance can fall to either side.

Pi vs Tau: Pi's Rebuttal

As you may know, I have been strongly intrigued by the tau movement. This movement is promoting the idea that we should not be using pi as the circle constant, but rather 2π, which has been renamed with the greek letter tau.

This movement was started by Bob Palais of University of Utah when he wrote the article Pi is Wrong which was published in the Mathematics Intelligencer. Physicist Michael Hartl proceeded to write The Tau Manifesto, and founded Tau Day on June 28th (instead of Pi Day on March 14th) as well as the website www.tauday.com. This gave dozens of reasons in geometry, trigonometry, physics, and statistics why tau is more practical and natural than pi. On a side note, I currently have the world record for tau memorization at 2012 digits, but that's nowhere near the world record for pi memorization at 67890 digits.

If you have some more time, The Tau Manifesto is a fascinating read. I'd also recommend watching this video done by Vi Hart for an overview of the tau movement.

You can also click here to see all of the blog posts I have done giving details on why tau is better than pi.

Just two days before writing this, I found that someone was inspired by Hartl's Tau Manifesto to write The Pi Manifesto, which presents an interesting rebuttal for tau. This wasn't recent news in mathematics, but it was definitely news for me!

I found that all of the reasons for either side can actually be debated. I think this is the first post where there is actually a debate over what is the right answer.

An argument for tau is that the radius is what a circle is measured by. Dictionary.com defines circle as a closed plane curve consisting of all points at a given distance from a point within it called the center. Clearly, the radius is the main measurement here.

Pi supporters would then argue that the radius can only be found by taking the diameter and dividing it by two. When looking at a circle on paper, it is impossible to pinpoint the center and find the measurement out to the end.

Yet, constructing the circle requires knowing the radius. If you were to use a compass, you must put the point where you want the center to be and trace a line with a constant distance around it. This is the true way to form a circle. For practical uses, you would need circles to see what restaurants were within ten miles or something. In this case, ten miles is the radius, and you are constructing a circle with this measurement.

This example requires the choice between constructing the circle easier or deconstructing the circle easier. The construction requires the radius, while a circle already given uses the diameter.

An argument for pi is that the area formula, which is one of pi's most common uses, is messed up by this change.

A = πr^2  –>  A = 1/2τr^2

Yet, tau-ists argue that this is more natural. First, there are many other shapes that have 1/2 at the beginning of their area formula. Triangles, trapezoids, and n-sided polygons all use formulas with a 1/2 in it. Second, the area formula can be proven (I will in a future blog post), and this proof's last step is to multiply 2πr^2 and 1/2. It turns out to be more natural just to tack on the 1/2 rather than hiding the 2π with it.

This is again an argument where we need to choose between having a natural number or an efficient number. It is clear that tau is the constant that belongs, but we have an opportunity to simplify the equation.

Both manifestos list other reasons that take longer to explain, but are very interesting (I think a lot more interesting than the two above). These involve finding measurements of the unit circle, graphing trigonometric functions, and rewriting Euler's identity.

Because math is such a definitive discipline, it is rare for Cool Math Stuff posts to have comments. I strongly encourage you to comment on this post. After reading both manifestos, watching several YouTube videos, and seeing lots of press coverage, it is difficult to take a side. Comment what you think about the different arguments (I think the three I listed earlier are the closest arguments to call), and if you are a Tau-ist or not. This is one of the few chances where you can get into a debate about mathematics!

Central Angle Theorem

Because it is difficult to do on the computer, I rarely post geometry-related posts. However, I think this one is worth the extra time. While I go through this proof, I strongly suggest that you grab some scrap paper and draw the circles along with me. It will make much more sense this way.

Look at the circle below. I randomly placed points A, B, and C, and the center is marked O. I also marked the angle ACB.

First of all, if you move the point C to wherever you want on the circle, the angle ACB will still have the same measure. The reason of this will become clear after I prove the central angle theorem.

The central angle theorem states that angle AOB is always double of ACB, regardless of the placement of the points A, B, and C.

The proof of this will take three steps. For the first step, I have moved the point C so that the line AC goes through O. Basically, AC is the diameter of the circle.

I have marked the angle ACB with a thick line and angle AOB with a thin line. There is also a dotted line to further identify angles.

First, we must take note of the fact that since O is the center of the circle, any line containing O in it is a radius of the circle. So, the line AO, BO, and CO are all radii of the circle. Since the radius is always the same measurement, AO, BO, and CO are congruent. The tick marks on these three lines are a way of saying that all lines with that mark are congruent.

Now, we are going to use some triangle theorems. I don't have time to prove them both now, but I'm sure you could prove them pretty easily.

Isosceles Triangle Theorem: If two sides of a triangle are congruent, their opposite angles are congruent. In this example, since BO and OC are congruent, the angles BCO and OBC are congruent.

Exterior Angle Theorem: If you have a triangle where an exterior angle is shown, its measurement is the sum of the two remote interior angles. For instance, in the triangle below, angle d is the sum of angle a and angle c. In the example above, angle AOB is the sum of angles BCO and OBC.

Because of the isosceles angle theorem, we already determined that angle BCO is congruent to angle OBC. Rather than just marking them congruent, let's let x be their measurement.

Now, we will use the exterior angle theorem. We already determined that angle AOB is the sum of angles BCO and OBC. And, we already determined that these two angles have a measure of x. So, we can determine the measure of angle AOB.

m<AOB = m<BCO + m<OBC

m<AOB = x + x

m<AOB = 2x

So, angle AOB has a measure of 2x.

The central angle theorem says that the measure of angle AOB is double of the measure of angle ACB. As you can see, m<AOB = 2x and m<ACB = x, so we have proved the first part.

Now, imagine that C was not on the diameter. Say you had an image like the following:

In this example, I have kept the information that we already found. The angles that are thick lines are the ones we are testing. Before we continue, we must first identify a few angle measurements. All of them will be found using the same theorems as before. I have made the angle measures blue so it is easier to keep track.

Also, note the fact that a triangle's angles always add up to 180°.

If you look at triangle AOB, you will see that all three angles have a measurement. We know that if we add these up, we will have to get 180.

2x + y + y = 180

2x + 2y = 180

This fact will definitely help in the future. Also, we should look at the fact that BO and OC both have a tick mark, meaning that they are congruent. Because of the isosceles triangle theorem, we know that <BCO and <OBC are congruent.

In triangle BOC, we have a few of the angle measurements. One of them is 2y, which is already a part of the 180° in the triangle. We also have a 2z, which is not already a part of it. So, we will must add 2z to the equation. But, when you add 2z, you must also subtract back out a 2z. So, it will look like this:

2x + 2y = 180

2x + 2y + 2z - 2z = 180

2y + 2z + 2x - 2z = 180

(2y + 2z) + (2x - 2z) = 180

We have now separated what we know already from the new information, which is that the rest of the angles add up to 2x - 2z. But, we also determined that they are congruent. So, we will just divide 2x - 2z by two to find out these angle measurements.

(2x - 2z) = 2(x - z)

So, this measurement is x - z.

But, this proof requires the measure of angle ACB. We almost have it. We found that it is made up of z and x-z.

z + x - z = x

So, m<ACB = x. Again, this is half of 2x, so we have proven this type of example.

We have proven it if C is on the diameter or past the diameter. But, what if it is before the diameter?

I have again labeled some angle measures using the isosceles triangle theorem and exterior angle theorem. This time, the new variable is n. Note that as before, 2x + 2y = 180.

Angle BAO has a measure of y and angle CAO has a measure of n. So, angle BAC should therefore have a measure of y-n.

So, let's see what measurements we have for triangle ABC. We have the y-n, the y, and the x, as well as two unidentified angles that must sum to 2x + 2y.

y - n + y + x + ___ = 2x + 2y

2y + x - n + ___ = 2x + 2y

___ = x + n

So, we must figure out how to split up the x + n between angles CB• and ACB.

If you look at triangle BOC, you will notice that there are two congruent sides! So, we can use the isosceles triangle theorem to say that <OBC is congruent to <OCB. <OBC has a part with x and <OCB has a part with n.

In order to make them equal, we must give <OBC the n and <OCB the x. This will make both measures x+n, which will allow them to be congruent.

And where does this x go? x is put as the measure of <ACB. Since <AOB is double of this, we have proved the last possibility.

These theorems and variables are pretty complicated. However, like I said in the beginning, it will make sense if you actually try to prove it yourself. Draw a circle and start trying to prove the different possibilities. This is how I was able to understand it. I hope this is a guide to try to prove it, but it may be difficult to understand without actually following along.