## Saturday, February 23, 2013

### The Origins of "Eureka!"

Possibly one of the greatest mathematicians of all time is Archimedes. He was born in 287 BC in the Greek colony of Syracuse. Archimedes advanced geometry, founded integral calculus, and approximated the value of pi.

King Hieron had a goldsmith create him a gold crown, using the cold the king gave him. When the crown was returned, it weighed the same amount as the gold he had to work with. However, the king was suspicious as to the material used in the crown.

He called upon Archimedes to investigate the problem, who did not know what to do at first. However, he had an epiphany while sitting in a public bathtub. He realized that when he sat in the water, the water level rose. After discovering these concepts of water displacement, he was ecstatic.

He was so excited that he jumped out of the bathtub, and without putting on any clothing, ran through the crowded streets shouting "Eureka!" (I have found it!) The crown was later tested and it was made of silver.

Now, the term "Eureka" is used in many instances. For instance, the California state motto is "Eureka," referring to the fact that people found gold there in 1848. In fact, there is even a town named Eureka.

I found it interesting that this exclamation dates back to one of our most famous mathematicians and their extremely over the top reaction.

## Saturday, February 16, 2013

### Handshake Problem

Let me start by giving a little math problem. n people are in a room, and each one has to shake every other person's hand. How many handshakes will it take for all n people to have shaken every other hand?

Let's first try a few numbers. For two people, it would obviously just take one handshake. For three, it would take three handshakes. For four, it would take six handshakes. For five, it would take ten handshakes.

Do you see the pattern? 1, 3, 6, and 10 are the triangular numbers. In fact, this pattern always continues.

If you look at it logically, you will see why. Person 1 has a number of hands to shake. Person 2 would have to shake all of the hands except for person 1's (it was already counted). Each person has to shake one less than the one before until there is just one left.

Adding up these handshakes will be a sum of the first n natural numbers, which is the definition of triangular numbers.

I found it cool that a famous number sequence could be applied to a practical problem like this.

## Saturday, February 9, 2013

### 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!

## Saturday, February 2, 2013

### 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.

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.

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