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October 9th, 2017, 07:24 PM  #1 
Member Joined: Apr 2017 From: Canada Posts: 32 Thanks: 2  Pythagoras Theorem in 3D
I was wondering if you have a 3D triangle, sort of like a wedge, could you use 2 of the wedge's sides to find the third? From Fermat's Last Theorem, we know that $\displaystyle a^3 + b^3 ≠ c^3$ However, is there ANY pattern or relationship between the legs and the hypotenuse? In other words, is there a general way to use $\displaystyle a,b$ to find $\displaystyle c$? 
October 9th, 2017, 07:44 PM  #2 
Senior Member Joined: Aug 2012 Posts: 2,007 Thanks: 574 
The diagonal of a 3D rectangle  a rectangular parallelepiped  is given by $L = \sqrt{l^2 + w^2 + h^2}$; and you can derive that by using Pythagoras in any two dimensions and then in the third. To illustrate with an example, what is the diagonal of a unit cube? The diagonal of any 2D face is of course $\sqrt{2}$ by Pythagoras. Then the diagonal of the cube is the diagonal of a 2D triangle with legs $1$ and $\sqrt 2$, namely $\sqrt{1^2 + (\sqrt{2})^2} = \sqrt{3} = \sqrt{1^2 + 1^2 + 1^2}$. This idea generalizes to the formula for the length of a vector in $n$dimensional Euclidean space, $\displaystyle L = \sqrt{\sum_{i=1}^n x_i^2}$ where the $x_i$'s are the coordinates of the vector. Last edited by Maschke; October 9th, 2017 at 08:15 PM. 
October 10th, 2017, 09:27 AM  #3  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894  Quote:
Quote:
 
October 13th, 2017, 09:32 AM  #4  
Member Joined: Apr 2017 From: Canada Posts: 32 Thanks: 2  Quote:
 
October 13th, 2017, 09:37 AM  #5 
Senior Member Joined: Aug 2012 Posts: 2,007 Thanks: 574  Just take the nth root of a^n + b^n. It won't be an integer in general but you can certainly find c. For example 3^4 + 5^4 = 81 + 625 = 706, and the 4th root of 706 is 5.1546...

October 14th, 2017, 04:01 AM  #6  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894  Quote:
Quote:
 
October 14th, 2017, 05:09 AM  #7  
Senior Member Joined: Jun 2015 From: England Posts: 853 Thanks: 258  Quote:
So the sentence might be more accurately phrased "bytaking an nth root....."  
October 15th, 2017, 12:14 PM  #8 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
Yes, and you should have said "Possibly it is worth making the distinction between the nth root function, which is single valued by definition and an nth root itself which may be one of several values.

October 15th, 2017, 12:17 PM  #9 
Senior Member Joined: Aug 2012 Posts: 2,007 Thanks: 574 
Beg to differ w/Country Boy and studiot. OP specified triangles. We're talking about positive numbers. The nth root is unambiguous. Last edited by Maschke; October 15th, 2017 at 12:22 PM. 

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