Had a great time this morning with my older son talking about right triangles. The section we were in today discusses properties of 45-45-90 and 30-60-90 triangles. He was quite surprised to see the factor of arise in the 30-60-90 triangle (and following that in the calculation of the area of an equilateral triangle).
I wanted to expand on that surprise a little and was going to mention that you see in a pentagon:
Me: “Well, you see in a square . . . .”
him: “I wonder if you see in a cube.”
Oh yes, that’s a much better thing to talk about than seeing in a pentagon. We used a Rubik’s cube and our Zometool set to see if we could get to the bottom of this idea. We started off the talk with the square as a reminder of where the comes from.
Next we took a look at a cube to see if we could find any ‘s hiding anywhere:
Now, just for fun, what about a hyper cube? I wish I had better drawing skills, but at least this shows the idea. I did not know that the long diagonal of a hyper cube forms a 30-60-90 triangle, so that was a neat surprise connection with what we were taking about today:
So, a fun and unexpected little project about square roots. Probably would be neat to revisit something like this whenever we start talking about the distance formula since this project essentially shows how you calculate distances in dimensions higher than 3. Nice morning.
3 thoughts on “Did you know that there is a 30-60-90 triangle in a hyper cube?”
I wondered about the long diagonal of a cube myself, because it does contain the square root of 3. However, just because a length involves the square root of 3 doesn’t make it a 30-60-90. The ratio of the sides of a 30-60-90 is
x: x*(square root of 3): 2x
Where x is opposite the 30 degree angle, x*sqr root of 3 is opposite the 60 degree angle and 2x is opposite the right angle.
The long diagonal’s ratio is:
x: x*sqr root of 2: sqr root of 3
In the long diagonal, x is the length of an edge of the cube, x*sqr root of 2 is the diagonal of a face of the cube, and x*sqr root of 3 is the length of the long diagonal of the cube. Therefore this is not a 30-60-90 triangle.
Using trigonometry I calculated the angles. Looking at the smallest angle, opposite over hypotenuse is 1/(sqr root of 3) which is 0.57735. The arcsin of 0.57735 is 35.26 degrees. The other non-right angle has an opposite over hypotenuse ratio of (sqr root of 2)/(sqr root of 3) which is 0.8165. The arcsin of 0.8165 is 54.74 degrees.
In a 30-60-90 triangle, the ratio for the smallest side to the hypotenuse = x:2x which equals 0.5, yielding an arcsin of 30 degrees. The final ratio for opposite over hypotenuse in the 30-60-90 gives us a ratio of x*(sqr root of 3) / 2x which simplifies to (sqr root of 3)/2 or 0.86602. Sure enough, the arcsin of 0.86602 is 60 degrees.
The takeaway for me is that we should not be fooled into thinking we have a special right triangle just because we see familiar lengths that involve sqr root of 2 or the sqr root of 3. The only reason that the super-diagonal of the cube has the square root of 3 in it is because when you define the edge length of the cube as x, the diagonal of the face of the cube is x*(sqr root of 2) and because c-squared = a squared plus b-squared you end up with:
c-squared = (x * x) + [x*(square root of 2) * x*(square root of 2)]
c-squared = x-squared + [2 * x-squared]
c-squared = 3*(x-squared)
Take the square root of both sides:
c = x*(square root of 3) which is the formula for the long diagonal of the cube where x is the length of the edge of a cube and c is the long diagonal.
Thanks and peace out
Correction: Introduction should read The triangle containing the long diagonal has sides in the ratio:
x: x*sqr root of 2: x*sqr root of 3
My post is about a 4 dimensional cube rather than a 3 dimensional cube.
You are correct that the triangle in a 3 dimensional cube is not a 30-60-90 triangle.