A Sierpinski Triangle Problem

Mathematics is a strange and mysterious thing. It is as if it is almost magical, able to do things that our physical beings cannot percieve. Perhaps a very good example of this is a problem that I encountered just yesterday, having to do with the Sierpinski Triangle.

First of all, you might ask, what in the world is a Sierpinski triangle? Well, it is a fractal, in which it creates a self-similar pattern within itself. Perhaps a picture is better-

As you can see, the pattern here is just forming triangles inside every dark triangle on and on and on. In fact, this pattern can go on for infinity, as it will in the problem I will be showcasing below:

The segments joining the midpoints of the sides of an equilateral triangle are drawn, and the interior of the triangle they form is removed from the interior of the original triangle. The segments connecting the midpoint of the remaining triangles are joined, and the interiors of the triangles they form are removed from the interior of the original triangle. If this process is repeated forever, how much of the orginial interior will be left?

Let’s take a look back at the picture I showed you above. Suppose the black spaces were the interior of the triangle. If we were to remove the triangles, they would then leave white triangles behind. In other words, the black spaces are the interior, and the white spaces are the places where triangles have been removed. Now let’s go back to the problem. What are we trying to find? Well, we are trying to find out how much of the original interior (aka how much of the first triangle in the picture above) is left. So how do we approach this?

Perhaps we can find the total amount of area that was removed from the original triangle, or in other words, the total area of white space. We would then get that and subtract it from the area of the original triangle. This difference (aka the amount of black space left after all the removing) would then be divided by the area of the original triangle to get a fraction of how much is left of the original triangle. That’s our process.

Let’s begin. First thing- how do we calculate the total amount of space removed? In other words, how do we calculate the total amount of white space? Well,  let’s take this as an infinite geometric sequence. In our first cut, ¼x is removed, where x is the original triangle’s area. This is indicated by the second triangle in the picture above.  In our second cut, 3 (1/16) x is removed, as indicated by the third triangle above. In our third cut, 9 (1/64) x is removed, indicated by the fourth triangle. This pattern then continues on infinitely. Again, it is a sequence- (1/4)x, 3(1/16)x, 9(1/64)x,…… So in order to find the total amount of white space, we have to add all the terms here up. You might ask, well this is infinite, so how in the world do we add this up? Note that the first term is (1/4)x and the ratio is 3/4. We now use the formula for an infinite geometric series:

where a₁ is the first term, and r is the common ratio. Both of these we know. So let’s substitute. S= [(1/4)x] / [(1- 3/4)] = [(1/4)x] / [1/4] = x. Thus, a total of x is removed from the triangle.

But wait, isn’t x the area of the original triangle, too? This means that if a total of x is removed, then in essence, the whole triangle has disappeared. Thus, the answer is that zero percent of the original interior is left. Everything has disappeared. Now picture this in your head. You are given a paper triangle. You are to cut out a triangle whose edges are the midpoints of the original triangle. You do this for every triangle that you see after you cut every time. And not only that, this is repeated infinitely. If so, then in the end, you get nothing.

To me, that’s really hard to imagine. How in the world do you get nothing if you keep on cutting a paper? It’s crazy and bizarre. Yet, at the same time, this is what makes math beautiful. And in this case, seemingly magical.

A Mathematical Beauty of Hyperbolas

This morning in my Algebra 2 Honors class, my teacher decided to show my class and I some extra tid-bits. These tid-bits were perhaps one of the most beautiful mathematical concepts I ever saw. We were learning about analytic geometry and hyperbolas at that time. For those of you who don’t know what a hyperbola is, check this graph below:

To sum it up in general terms, a hyperbola is pretty much any graph that looks like the one above- the red curves only-(vertical or horizontal). However, there is one property of the hyperbolas that I will focus on for this post: notice that besides red curves, there are black lines also. These black lines are asymptotes, in which the graph of the hyperbola- the red curves- approach them but never ever touch them.

Visually, yes, we understand this property. However, what my math teacher showed me this morning was how to explain this in mathematical terms, specifically in limits. Isn’t that almost magical? The fact that you can explain a visual quality in numbers and equations?

Anyway, let us first take take the equation of this hyperbola in the graph above.

 x²/16 – y²/9 = 1

Notice that in order to illustrate this relationship in which the graph approaches but never touches the asymptotes, y has to be near positive/negative infinity. So let us get the y-coordinates of this equation. To do so, simply solve y in terms of x using algebra. This will result in-

                                         y=±¾ √(x²-16)

   [which equals by factoring out an x² in the radical..] y=±¾ √[x²(1 –  16/x²)]

   [which equals by taking the x^2 out of the radical..] y=±¾x √(1 –  16/x²)

 Notice again that for this relationship to occur not only does y have to near pos/neg infinity, so does x. So suppose x is approaching negative or positive infinity. Or in math, x→±∞. Let us using this value plug x into the equation that we have now. This will cause 16/x² to be approximately equal to 0, because 16, a small number, out of infinity, a super large number, is a really small fraction that has to be very very close to zero. In math terms, 16/x²→0.

Now, let us get this value for 16/x^2 and plug it back into the equation. This will cause the terms under the radical to go like this:

Since 16/x²→0, (1 –  16/x²)→ (1-0) → 1

Substitute this (1 –  16/x²)→ 1 into the equation y=±¾x √(1 –  16/x²) and get-

y→±¾x √(1 –  16/x²)→ ±¾x √(1)→ ±¾x

given x→±∞ ; y→±¾x

Tada! Let’s look at this limit equation we got. First of all, refer back to the graph. Notice that the ±¾x  is in fact the equations of the asymptote lines. So what this limit equation is basically saying is that as x approaches negative/positive infinity, y or the function will approach  ±¾x aka the asymptotes. In other words, we have just mathematically illustrated the asymptotic relationship in which the hyperbola will near but never touch the asymptotes. Not just that, we in a sense derived it from a hyperbolic equation; thus we proved that in any hyperbola, the graph will always have this relationship. Isn’t that beautiful?

Perhaps the way I’m showing it to you on the computer is not as beautiful as it really is. (It’s actually pretty hard to write these equations on a computer. I couldn’t even use Words for this.) I would agree that showing this on paper and actually writing this out makes this derivation seem so much more beautiful.

But what makes this beautiful, computer or not? I dunno really, but I think it is the fact that it is so simple and so elegant. Just a small number of steps and you can explain a visual property in mathematical terms. I don’t know about you, but whenever I read through this process, there is a thrill down my spine. It is just so mathematically beautiful.

Anamorphism in Art

Hello guys. Today I will be switching my view from Jerusalem to art, specifically illusion art. First check out this video below:

Pretty cool, isn’t it? This in art science is called anamorphism. Anamorphism is when a projection of an object is distorted, but when you are at a certain angle, the image is reconstituted. For example, in the video above, the Rubik cube on paper is a planar projection of the actual Rubik cube. You first begin looking at this projection which makes it seem as if the actual Rubik cube was there. This is the point of view when the image is reconstituted. However, as the projection is turned around, it doesn’t seem like the actual Rubik cube anymore.

What you just saw is perspectival anamorphism. A cooler type is mirror anamorphism. It’s hard to describe this, but it is pretty much the same concepts with the exception of having to involve a mirror. Check out this video to see what I mean:

Well, of course, instead of just saying ooh and aah, I wanted to know the science behind it. However, I discovered it was not science, but geometry that was behind all this. Check this webpage: Descriptive Geometry and Anamorphis (link).

Well, hope it’s interesting. That’s all for today.

Why C^2 = A^2 + B^2

One of the most basic formulas in geometry is the Pythagorean theorem. It is mistakenly understood that Pythagoras, a famous Greek mathematician, was the one who found it, but ancient papers have shown that the Chinese had proved this theorem first. Nevertheless, no matter who proved it first, there have been many proofs of the Pythagorean Theorem, all which emphasize the beauty of this formula.

         Pythagoras

Pythagorean Theorem

Check out this webpage on the most common proofs of c^2=a^2+b^2: Pythagorean Theorem Proofs