
SOMA News 
17 september 2020
EMail. 
Interesting SOMA height and Mathematics.
By 20200926 Newsletter released on the web. 
By 20200927 Additional Mathematics from Bob Nungester. 
Today 20200917, my friend Tim Fielding =/\= wrote:
Interesting that piece #1 (v) is level, even though piece #2 (L) isn't, hmmmmm. Try it out. Have fun...
Now this is interesting indeed. Is it really level, or is it just approximate. ??
Well, let's look at the math behind this piece placement.
Doing so involve the calculation of sine and cosine to an angle. So I define a small triangle (drawing A)
The hypotenuse has the length of "1", being equal to the side length of one cube. (colored Blue.)
The two catheters on the triangle will then have a length determined by cosine and sine.
cos(v) is colored Red, and sin(v) is colored Green.
They will turn into numbers when we calculate the angle v, which is here colored Pink.
SO: Now to the mathematics.
We know that the L piece rests on a single cube. (drawing B)
So looking at the subdrawing (drawing C). We see that we "go up" by two Green, and then "down" by one Red. And then we are at height 1.
Using math, this is sin(v) + sin(v)  cos(v) = 1
We solve this 2 sin(v)  cos(v) = 1 ⇒ v = 53,13037°
Now we have the angle v, and that allow us to compute the height of piece L in this position.
Looking at the partial figure (drawing D), we find the height to be: 3 Green + 1 Red
H = 3 * sin(v) + cos(v)
Which, knowing the angle give H = 3 * sin(53,13037) + cos(53,13037)
H = 3 * 0,8000028 + 0,5999962 giving
H = 3,0000047
Which basically is exactly 3 cube heights, to a resolution of approximately 5 parts in a million
So, now we know  that an L piece resting on a single cube, will have its top edge at EXACTLY 3 cube heights.
Wow, what a beautiful coincidence.
Thank you Tim for discovering this pretty geometric construction.
A simpler and more precise Math.
20200927 Ok, that was quick.
Only a few hours after my release of this page (at 20200926), my friend Bob Nungester reacted.
Sending back the most beautiful proof of the Mathematics. This only proves how one person (me) can be so focused on one solution, that it blocks for a more obvious way.
I have, just last week, been teaching my second graders how to find the derivative of trigonometric functions. [ f(x)=sin(x) ⇒ f '(x)=cos(x) ] My focus was thus on trigonometrics.
But let's hear what Bob is saying:
Tim's newsletter is indeed very interesting.
I wondered if it could be proven to be exactly equal (no rounding errors) and developed this solution:
Call the Green length X and the Red length Y. Then our equation is 2*X  Y = 1
We also know from the Pythagorean theorem that X^{2} + Y^{2} = Z^{2}. In our case X^{2} + Y^{2} = 1
So here is the mathematics.
 Our equation is   2*X  Y = 1  
(1)  Rearranging   Y = 2*X  1  
 Substituting Y into Pythagoras   X^{2} + (2*X  1)^{2} = 1  
 Expanding the parenthesis   X^{2} + (4*X^{2}  4*X + 1) = 1  
 Reducing   5*X^{2}  4*X = 0  
 Dividing all by X [note 0/X = 0]   5*X  4 = 0  
(2)  Rearranging for X   X = 4/5  
    
 Now the height of piece #2    
(H)  The height of piece #2 is   H = 3*X + Y  
 Substituting Y from (1)   H = 3*X + (2*X  1)  
 This reduces to   H = 5*X  1  
 Inserting X from (2)   H = 5*(4/5)  1 ⇒  H = 3 
So the height is indeed exactly 3 cubes. Proven; using only integer coefficients in the equations.
Thanks to Tim for a very interesting problem.
And thank you to Bob for an excellent and simple proof.
Made by Tim Fielding <macteabird@gmail.com>
Mathematics and editing by Thorleif Bundgaard <thorleif@fambundgaard.dk>
Improved mathematics by Bob Nungester <bnungester@comcast.net>
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