Contained herein: Method for finding the height and center of an equilateral triangle and a tetrahedron in terms of side length. Two methods to find the angle formed by the center and any two vertices, AKA the tetrahedral angle. A python program for calculating the above. How to make a paper tetrahedron.
This proof involves a basic understanding of trigonometry and a good understanding of algebra. Skip down to the second proof it this looks too complex.
Since all sides and angles are equal all the heights are the same and intersect at the center which is the same distance from each side. This allows us to set up this Pythagorean identity and solve for x.
$\begin{array}{c}{(h-x)}^{2}={x}^{2}+{\left(\frac{a}{2}\right)}^{2}\Rightarrow {h}^{2}-\mathrm{2hx}=\frac{{a}^{2}}{4}\Rightarrow -\mathrm{2hx}=-{h}^{2}+\frac{{a}^{2}}{4}\Rightarrow x=\frac{h}{2}-\frac{{a}^{2}}{\mathrm{8h}}\Rightarrow \\ x=\frac{\sqrt{{a}^{2}-\frac{{a}^{2}}{4}}}{2}-\frac{{a}^{2}}{8\sqrt{{a}^{2}-\frac{{a}^{2}}{4}}}\Rightarrow x=\frac{4\left({a}^{2}-\frac{{a}^{2}}{4}\right)-{a}^{2}}{8\sqrt{{a}^{2}-\frac{{a}^{2}}{4}}}\Rightarrow x=\frac{{\mathrm{4a}}^{2}-{a}^{2}-{a}^{2}}{8\sqrt{{a}^{2}-\frac{{a}^{2}}{4}}}\Rightarrow \\ x=\frac{{a}^{2}}{4\sqrt{{a}^{2}-\frac{{a}^{2}}{4}}}\end{array}$The height of of a face, the x of an adjacent face and the height of the tetrahedron form a right triangle. since we already know h and x we can find H with the Pythagorean theorem.
$\begin{array}{c}{x}^{2}+{H}^{2}={h}^{2}\Rightarrow H=\sqrt{{h}^{2}-{x}^{2}}\Rightarrow H=\sqrt{{\sqrt{{a}^{2}-\frac{{a}^{2}}{4}}}^{2}-{\left[\frac{{a}^{2}}{4\sqrt{{a}^{2}-\frac{{a}^{2}}{4}}}\right]}^{2}}\Rightarrow \\ H=\sqrt{{a}^{2}-\frac{{a}^{2}}{4}-\frac{{a}^{4}}{16\left({a}^{2}-\frac{{a}^{2}}{4}\right)}}\Rightarrow H=\sqrt{{a}^{2}-\frac{{a}^{2}}{4}-\frac{{a}^{4}}{{\mathrm{16a}}^{2}-{\mathrm{4a}}^{2}}}\Rightarrow H=\sqrt{{a}^{2}-\frac{{a}^{2}}{4}-\frac{{a}^{2}}{12}}\Rightarrow \\ H=\sqrt{\frac{{\mathrm{12a}}^{2}-{\mathrm{3a}}^{2}-{a}^{2}}{12}}\Rightarrow H=\sqrt{\frac{{\mathrm{2a}}^{2}}{3}}\end{array}$This is the 3D version of finding the center of the equilateral triangle. The two heights of the tetrahedron are the same, go from the center of a face to the opposite vertex, and intersect at the center. Since we know H, h, and x we can solve for the triangle y, H-y, and h-x.
$\begin{array}{c}{(H-y)}^{2}={y}^{2}+{(h-x)}^{2}\Rightarrow {H}^{2}-\mathrm{2Hy}+{y}^{2}={y}^{2}+{(h-x)}^{2}\Rightarrow -\mathrm{2Hy}={(h-x)}^{2}-{H}^{2}\Rightarrow \\ y=\frac{H}{2}-\frac{{(h-x)}^{2}}{\mathrm{2H}}\Rightarrow y=\frac{\sqrt{\frac{{\mathrm{2a}}^{2}}{3}}}{2}-\frac{{\left[\sqrt{{a}^{2}-\frac{{a}^{2}}{4}}-\frac{{a}^{2}}{4\sqrt{{a}^{2}-\frac{{a}^{2}}{4}}}\right]}^{2}}{2\sqrt{\frac{{\mathrm{2a}}^{2}}{3}}}\Rightarrow y=\frac{\frac{{\mathrm{2a}}^{2}}{3}-{\left[\frac{4\left({a}^{2}-\frac{{a}^{2}}{4}\right)-{a}^{2}}{4\sqrt{{a}^{2}-\frac{{a}^{2}}{4}}}\right]}^{2}}{2\sqrt{\frac{{\mathrm{2a}}^{2}}{3}}}\Rightarrow \\ y=\frac{\frac{{\mathrm{2a}}^{2}}{3}-{\left[\frac{{\mathrm{4a}}^{2}-{a}^{2}-{a}^{2}}{4\sqrt{{a}^{2}-\frac{{a}^{2}}{4}}}\right]}^{2}}{2\sqrt{\frac{{\mathrm{2a}}^{2}}{3}}}\Rightarrow y=\frac{\frac{{\mathrm{2a}}^{2}}{3}-\frac{{\mathrm{4a}}^{4}}{16\left({a}^{2}-\frac{{a}^{2}}{4}\right)}}{2\sqrt{\frac{{\mathrm{2a}}^{2}}{3}}}\Rightarrow y=\frac{\frac{{\mathrm{2a}}^{2}}{3}-\frac{{a}^{4}}{{\mathrm{4a}}^{2}-{a}^{2}}}{2\sqrt{\frac{{\mathrm{2a}}^{2}}{3}}}\Rightarrow \\ y=\frac{\frac{{\mathrm{2a}}^{2}-{a}^{2}}{3}}{2\sqrt{\frac{{\mathrm{2a}}^{2}}{3}}}\Rightarrow y=\frac{{a}^{2}}{6\sqrt{\frac{{\mathrm{2a}}^{2}}{3}}}\end{array}$From the above we can see the proportions of the triangle y:h-x:H-y is 1:2sqrt(2):3
By the time I was finished with all the above algebra I was tired of doing math but I needed to check my work. The solution was the following python program. It's written in python v2.7. Unlike many languages, ** is python's exponent operator
import math
def h(a): #finds the hight of equelateral triangle
return math.sqrt(a**2-a**2/4.)
def x(a): #finds the distance from the side to the center of equelateral triangle
return a**2/(4*h(a))
def H(a): #finds the hight of tetrahedron
return math.sqrt((2*a**2)/3.)
def y(a): #finds the distance from any face to the center of tetrahedron.
return a**2/(6*H(a))
def complementAngle(a): #finds angle of vertice-tetraCenter face-tetraCenter lines
return math.degrees(math.asin((h(a)-x(a))/(H(a)-y(a))))
def angle(a): #finds angle of vertices from center. YAY!
return 180-complementAngle(a)
#test against values that I found doing the algebra with 2 instead of a.
print "h(2)",h(2), math.sqrt(3)
print "x(2)",x(2), math.sqrt(1./3)
print "H(2)",H(2), math.sqrt(8./3)
print "y(2)",y(2), 2/(3*math.sqrt(8./3))
# test to show the numbers are consistent with themselves. It's the Pythagorean theorem.
print y(2)**2+(h(2)-x(2))**2,(H(2)-y(2))**2
#test to show that it works with any int or float.
print
for a in range(1,9):
print angle(a+.5*a%2),
This proof is significantly simpler than the algebraic proof and only requires the most rudimentary understanding of algebra, geometry, and trigonometry.
Start with an equilateral triangle of side length 2. Bisecting any of the angles will produce a triangle with angles 60:30:90 and sides1:2:sqrt(3). Bisect that 60 degree angle of the new triangle and you get a new 60:30:90 triangle. Since the two triangles are similar triangles we can find the two unknown sides from the one known side, length 1. The distance from the side to the center is 1/sqrt(3).
Since we know the hight of the equilateral triangle of any face is sqrt(3), and the hight of the center of any adjacent face is 1/sqrt(3), we can calculate the angle between faces as acos(1/3) which is approximately 70.53°
The cutaway view of a tetrahedron shows that the angle between two faces is the supplementary angle of the tetrahedral angle. 180°-70.53° = 109.47°. Yay!
Because my paper tetrahedron was instrumental in the development of my understanding of the pertinent geometry, I included this ViHart video about how to make one. Her blog is pretty cool too.
Some time ago I learned the bond angle for tetrahedral molecules is 109.5°. Thinking that was an exact value with some geometric significance, I set out to prove it. The first thing I did was make a paper tetrahedron. With that conceptual aid, some long bike rides, and a lot of algebra, I developed the included algebraic method. At some point I mentioned to my father what I was up to. He, in a surprisingly short amount of time, came up with the similar triangle solution using the arbitrary values. I think the message of this article is anybody can do math. I'm not a math major. I was taking my first calculus class when I did this. So, use your head.
Hey, this was really interesting. I just wanted to point out that a lot of your calculations for proof #1 could have been simplified intermediately. For example, the biggest source I see is when calculating y. You write (h-x)^2 = some mess, whereas we know from above that (h-x)^2 = x^2 + (a/2)^2. Also, sqrt(a^2 - .25a^2) is equivalent to sqrt(.75a^2) which is equivalent to a*root(.75) which also makes a lot of calculations easier.
Best,