Quite often, when doing calculations on polyhedra, you will find yourself with complex equations like $$\begin{equation}\tag{eq1}\label{eq1}\frac{20+7\sqrt3}{1+2\sqrt3}\end{equation}$$ Is the top evenly divisible by the bottom? I certainly can’t tell just by looking at them. There must be a method to factor the numerator. Let’s start with $$(A+B\sqrt{R})(C+D\sqrt{R})$$ multiplying the terms together gives $$AC+BRD+(BC+AD)\sqrt{R}$$ So, $$\frac{AC+BRD+(BC+AD)\sqrt{R}}{A+B\sqrt{R}} = C+D\sqrt{R}$$ Then we can substitute known values (A=1, B=2, R=3) into the formula. $$\frac{1C+2(3)D+(2C+1D)\sqrt{3}}{1+2\sqrt{3}} = C+D\sqrt{3}$$ Finding the answer is just a matter of finding C and D. Consider that $$\frac{\color{yellow}{C+6D} + \color{cyan}{(2C+D)}\sqrt3}{1+2\sqrt3} = \frac{\color{yellow}{20}+\color{cyan}{7}\sqrt3}{1+2\sqrt3}$$ We can break the equation into the colored sections to find $$\begin{align}\color{yellow}{C+6D} & \color{yellow}{=20} \\ C&=20-6D\tag{eq2}\label{eq2} \end{align}$$ and $$\color{cyan}{2C+D = 7}\tag{eq3}\label{eq3}$$ We now know C \eqref{eq2}, at least in terms of D, then substitute this into \eqref{eq3}. $$\begin{align}2(20-6D)+D &= 7 \\ 40-12D+D &= 7 \\ -11D &= 7-40 = -33 \\D &= \frac{-33}{-11} \\ D&=3\end{align}$$ Returning to \eqref{eq2} and substituting D=3, **…Read the Rest**

### sqr(A+B*sqr(C))

Lately, I have been doing a lot of math involving square roots of numbers added to square roots, in the form of \(\sqrt{A+B\sqrt{C}}\), this is called a “nested radical.” Normally, you would not be able to simplify any further, unless there was a common factor of both that could be removed, or if both items had the same number under the radical (eg. \(\sqrt{4\sqrt7+12\sqrt7}=\sqrt{16\sqrt7}=4\sqrt[4]{7}\) ). I can take a number like \((1+\sqrt5)^2\) and work out it is equal to \(6 + 2\sqrt5\). Unfortunately, it is very difficult to recognize that \(1+\sqrt5\) is the square root (or even a factor) of **…Read the Rest**

### SketchUp Platonics

I’ve been playing with Google’s Sketchup for a week or so and I had the thought of doing some polyhedra. It is often hard to mentally visualize a 3D object when looking at a 2D image, so I figured these would help. Here is the first SketchUp file, the five Platonic solids. You will need to install the program before you can view the file, but don’t worry, SketchUp is free. Each of the edge lengths is the same for all five objects, just to give good reference to size. In SketchUp, cubes are super-simple, so I purposely made it **…Read the Rest**

### Octahedron

The octahedron is a Platonic solid, which means it is made of all regular polygons for each face, being eight equilateral triangles arranged four at each vertex. With edge length of S, the surface area would be that of 8 equilateral triangles, \(8\cdot S^2\cdot \frac{\sqrt3}{4}=2\cdot S^2\sqrt3\). If we slice one in half, through 4 co-planar vertices, we get two 4-sided pyramids. Since we know that all the edges are the same length and all the angles are the same, the only possible shape for the base would have to be a square. If there were only three triangles meeting at **…Read the Rest**

### Law of Sines and Cosines

The law of cosines relates the sides and angles of a triangle. \(a^2=b^2+c^2-2bc\cdot \cos\alpha \\ b^2=a^2+c^2-2ac\cdot \cos\beta \\ c^2=a^2+b^2-2ab\cdot \cos\gamma\) It can also be rearranged to: \(\large\alpha=\arccos\left(\frac{b^2+c^2-a^2}{2bc}\right) \\ \large\beta=\arccos\left(\frac{a^2+c^2-b^2}{2ac}\right) \\ \large\gamma=\arccos\left(\frac{a^2+b^2-c^2}{2ab}\right)\) As long as all three sides or at least one side and two angles (which the third is trivial to find) are known, the remaining angles and sides are obtainable. You can also use the law of sines, \(\Large\frac{a}{sin\alpha}=\frac{b}{sin\beta}=\frac{c}{sin\gamma}\) or \(\Large\frac{sin\alpha}{a}=\frac{sin\beta}{b}=\frac{sin\gamma}{c}\), although I prefer the former as sin90°=1, therefore eliminating it in the fraction. Beware of ambiguous cases, such as if ∠α is acute and length a is shorter **…Read the Rest**

The cube is probably the most recognized and best known of and 3D shape. Kids young enough not to even know how to talk will still know about cubes, they play with multicolored wooden blocks. Older kids may tackle the Rubik’s Cube puzzle or play Yahtzee with dice. Then they may graduate to working in a cubicle as adults. We use the cube as a representation of the space around us. Pepsi was briefly sold in 24-can “cubes.” Apple’s G4 computer was cubic. Al Kaaba (literal English, the cube) is a cuboid shaped building in Mecca and the most sacred **…Read the Rest**

### Phi, the Golden Ratio

Phi \((\Phi, \phi)\) is a Greek letter that mathematicians have assigned to a specific ratio or proportion, called the golden ratio, that most people find to be attractive in art, architecture, and nature. The golden ratio is illustrated as \(\frac{A}{B}=\frac{A+B}{A}\equiv\phi\). The only positive solution is \(\displaystyle\phi=\frac{1+\sqrt{5}}{2}\)=1.6180339887498948482045868… Many items have this ratio embedded into their design, by choice or coincidence, such as; the Parthenon, credit cards, corporate logos, the Mona Lisa, and the layout of Quincy Park in Cambridge, MA. I find that there is some differences in the usage of symbols to represent the golden ratio. The most common is **…Read the Rest**

### Table of exact trigonometric functions

Since I noticed I had to keep looking some of these up, I place them here, just for reference, gathered from around the internet. Many of these formulas can be written in different ways, but I have simplified them as much as possible. Where, \(\phi=\frac{1+\sqrt{5}}{2}=1.6180339887498948482045868\ldots\), also known as the golden ratio (phi). a sin(a) = cos(b) = sin(90-b) = cos(90-a) tan(a) = cot(b) = tan(90-b) = cot(90-a) b deg rad deg rad 0 0 0 0 90 π/2 3 π/60 \(\frac{1}{16}\left[(2-2\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(-1+\sqrt{5})(1+\sqrt3)\right]\) \(\frac14\left[(2-\sqrt{3})(3+\sqrt{5})-2\right]\left[2-\sqrt{10-2\sqrt{5}}\right]\) 87 29π/30 6 π/30 \(\frac{1}{8} \left[\sqrt{30-6\sqrt5}-\sqrt5-1\right]\) \(\frac12\left[\sqrt{10-2\sqrt5}-\sqrt3(-1+\sqrt5)\right]\) 84 7π/15 9 π/20 \(\frac12\sqrt{2-\sqrt{2+\phi}}=\frac12\sqrt{2-\sqrt{\frac{5+\sqrt5}{2}}}\) \(1+\sqrt5-\sqrt{5+2\sqrt5}\) 81 9π/20 12 π/15 \(\frac{1}{8}\left[\sqrt{10+2\sqrt5}-\sqrt3(-1+\sqrt5)\right]\) **…Read the Rest**

### 22 miles straight up in 90 seconds

I may not involve geodesic shapes, but sure is awesome. http://hackaday.com/2011/10/10/22-miles-straight-up-in-90-seconds/ This is not a “because we can” moment, it is a “because nobody else can” moment. This is science at its finest. It hits Mach 3!!! Also see: http://ddeville.com/derek/Qu8k.html

### An unequal pyramid

Last post, I worked on the vertex edge angles of a triangular pyramid that had all equal angles originating from the apex. Using the same camera tripod analogy, take one of the legs (line DA) of the tripod and slide it out further from the other two, but keep it equally distant from each. You now have a camera that’s about to crash to the floor, but you also have a pyramid that has two equal apex angles and that leans. A right pyramid is a pyramid for which the apex lies directly above the centroid of the base. I **…Read the Rest**

### Tetrahedron

I started off posting about the truncated icosahedron, despite being fairly complex compared to the platonic solids, like the tetrahedron. The regular tetrahedron is probably the simplest 3D shape, except for maybe the sphere. It is made of 4 equilateral triangles, forming a triangular pyramid. Starting with a tetrahedron with vertexes named A, B, C, & D, with D as the apex (fig. 1), we can bisect ΔABC from point A to line BC and call that point G (fig. 2). Figure 1 Figure 2 Figure 3 The bisection of ΔDBC would follow the same, a line DG would also **…Read the Rest**