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Category Archives: Geometry
Casino Royale
I was re-watching Casino Royale recently and noticed some nice fractal imagery in the movie’s title sequence. I wonder what the Hausdorff Dimension of these fractals would be.
Smooth Curves
I’m not usually a fan of line graphs. I think connecting consecutive points on a scatter plot makes data that’s not continuous seem like it is. However, while plotting some data in Google docs earlier, a particular line graph did raise an interesting question in my mind.
Here are twelve consecutive attempts to solve a Rubik’s Cube and the time it took me for each in seconds. I think I’m doing well, though it’ll be some time before I get my times down below 60 seconds.
Here’s the same data, but with the points connected:
What got me thinking was the next one. Google docs lets you connect the points using smooth curves.
It’s very appealing visually, though somewhat inaccurate. Here’s what I’m wondering: how does Google docs draw the smooth curves?
Though there’s only one way to connect two points with a straight line, there seems to be infinite (or at least more than one) ways to connect two points with a curve. So which one is most appropriate? Sounds like I might have a project ahead of me involving Bézier curves.
The Tangent Identity in the Unit Circle
It’s been a fun couple days since learning about this tangent identity in a New York Math Circle class. First, I learned a few things about GeoGebra, and now after a few conversations over Twitter, I have a geometric interpretation of the equation
All the credit here goes to Colin Beveridge (@icecolbeveridge) of Flying Colours Maths. Here is the original picture he sent me.
If that makes sense to you, great! Otherwise, you can play around with the following GeoGebra applet (the blue points are the ones you can control), and I’m going to try to walk through the ideas below.
Instead of working with a non-right triangle, let’s equivalently consider three non-right angles A, B, and C whose sum is 180°. We can draw all three as adjacent central angles in a unit circle:
It is straightforward to say that together, the three angles subtend an arc of 180°.
Next, we construct three right triangles as follows. We draw a line perpendicular to one leg of angle A and tangent to the circle. We extend the other leg of angle A until these two lines meet to form a right triangle. This process is repeated for angles B and C.
Each blue triangle shown above has a base of 1 and a height equal to tan(A), tan(B), and tan(C) respectively. The combined area of the three blue right triangles is
Now we copy angle B to the position shown and extend the diameter of the circle to create another right triangle, this time in green.
The green triangle shares the tan(A) length with the blue triangle. The other leg of the green triangle has length tan(A)·tan(B) because:
Then we copy angle C and construct the yellow right triangle below.
Using a similar method, we can show that the longer leg of the yellow right triangle is equal to tan(A)·tan(B)·tan(C).
There is one more right triangle to construct. It’s shown in pink below.
The pink right triangle has a base of tan(A)·tan(B)·tan(C) and a height of 1. Therefore, its area is
But from our tangent identity, we know that the sum of the tangents of A, B, and C is equal to the product of the tangents and therefore the area of the pink triangle is equal to the sum of the areas of the blue triangles.
Wow.
The Tangent Identity in GeoGebra
The GeoGebra community is quite helpful. After posting about an interesting tangent identity yesterday, I set out to create a visual demonstration in GeoGebra. I had no idea where to begin.
I sent an email off and got a quick reply with a very informative screencast. The first version of the worksheet I made was a modified a bit by a helpful user (@kobak) and posted to GeoGebraTube. I went ahead and made another version suitable for the dimensions of this blog. You should be able to drag the vertices around to alter the angles of the triangle. The sum of the tangents and the product of the tangents remain equal!
Demonstration of the Tangent Identity
I’m normally not a big fan of Java, especially Java applets, but this is incredibly cool!
An Interesting Tangent
At last night’s session of the New York City Math Circle, I was introduced to the following theorem. Let A, B, and C be the angles of some non-right triangle. Then,
The proof is brief and involves use of the angle sum and difference formulas for tangent:
Here’s what I think is the much more interesting question: what in the world does this mean geometrically?
Equilateralness This Year
A couple months ago, good friend and colleague Mr. Honner asked a simple question motivated by the dates Oct 10, 2011 and Oct 11, 2011 (10-10-11 and 10-11-11). Which triangle is more equilateral?
In a follow up, he defined a new metric, “equilateralness”, which is given by the ratio of the triangle’s area to the area of an equilateral triangle with the same perimeter. As a triangle becomes more equilateral, this ratio approaches 1.
Today being another “close to equilateral date”, I decided to plot the equilateralness of the entire 2011 year. Obviously, we had one perfectly equilateral date (11-11-11), but did you know that today (12-12-11) is the second most equilateral at Eq ≈ 0.9953? And of dates that have proper triangle lengths January 11 and November 1 (1-11-11 and 11-1-11 respectively) share the lowest Eq this year?
You can explore the trends in the graph below (did you know you can publish charts from Google docs!?). If a date failed the triangle inequality (and there were 223 of them), I gave it an equilateralness of 0.
A “Colorful” Problem
Through the New York City Math Circle, I am taking a course for teachers called More Enrichment. Last night was the first class. I saw of a lot of great solutions, but what has me more excited are some great problems that remain unsolved. Here’s one that the presenter Gil Kessler filed under “Colorful Problems”.
Consider an 8 by 8 chess board:
and 31 dominoes:
If we try to tile the chess board with our 31 dominoes, we will leave two squares uncovered. Furthermore, of the two uncovered squares, one will be grey and one will be black.
The question is: if we arbitrarily remove two squares of different colors from the board, can we still tile the remaining squares with our 31 dominoes?
In class, it was mentioned that the answer is “yes”. Now to figure out the proof!
Tracing Orbits in the Mandelbrot Set
I recently added a new feature to my Mandelbrot Set script to trace the orbits of complex numbers through the plane. This along with the ability to zoom in, zoom out, or translate along the plane has made this a great teaching tool so far.
Some of my observations about the orbits:
- Points well outside the Mandelbrot Set are fairly mundane (as expected).
- Points along the edge of the main cardioid seem to offer the most visually appealing orbits (as shown above).
- Points close to but not inside the Mandelbrot Set show the most chaotic behavior.
I remember learning about the Mandelbrot Set in high school, but this is the first time I’m seeing this particular type of visualization. It’s very exciting!
A Golden Rectangle of Chocolate
Previously, I had wondered whether a Hershey’s chocolate bar approximated a golden rectangle. I will admit there’s an easier way to determine this (measure each side, divide, and see how close the ratio comes to φ), but why pass up an opportunity to play in GeoGebra or to demonstrate a golden rectangle construction?
First, I construct a square on the width of the chocolate bar.
Next, I connect point D to the midpoint of segment AC.
The green arc is part of a circle with center E and radius DE.
I extend AC until it intersects circle E at point F. I construct a line perpendicular to AF at F and extend BD. The rectangle outlined in red is a golden rectangle.
Looks like three-fourths of a Hershey’s bar is approximately a golden rectangle!