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October 9th, 2017, 07:24 PM   #1
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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 $\displaystyle c$?
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October 9th, 2017, 07:44 PM   #2
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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 2-D face is of course $\sqrt{2}$ by Pythagoras. Then the diagonal of the cube is the diagonal of a 2-D 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.
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Last edited by Maschke; October 9th, 2017 at 08:15 PM.
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October 10th, 2017, 09:27 AM   #3
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Quote:
Originally Posted by Antoniomathgini View Post
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 $\displaystyle a^3 + b^3 ≠ c^3$
Fermat's theorem says that there are no integer values of a, b, and c satisfying $\displaystyle a^3+b^3= c^3$. That has nothing at all to do with this problem, where a, b, and c can be any non-negative real numbers.

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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 $\displaystyle c$?
Yes, $\displaystyle c^2= a^2+ b^2$. The right angle part of a "wedge" is still two dimensional so lies in a plane.
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October 13th, 2017, 09:32 AM   #4
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Quote:
That has nothing at all to do with this problem, where a, b, and c can be any non-negative real numbers.
I apologize for phrasing my question incorrectly. Please forget anything I said about the wedge. What I meant to ask was whether or not there was any relationship between $\displaystyle a^n + b^n$ and $\displaystyle c^n$. We know that for any values greater than 2, $\displaystyle a^n + b^n ≠ c^n$. However, is there a general formula that would input $\displaystyle a^n + b^n$ and return/output $\displaystyle c^n$?
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October 13th, 2017, 09:37 AM   #5
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Quote:
Originally Posted by Antoniomathgini View Post
However, is there a general formula that would input $\displaystyle a^n + b^n$ and return/output $\displaystyle c^n$?
Just take the n-th 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...
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October 14th, 2017, 04:01 AM   #6
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Quote:
Originally Posted by Antoniomathgini View Post
I apologize for phrasing my question incorrectly. Please forget anything I said about the wedge. What I meant to ask was whether or not there was any relationship between $\displaystyle a^n + b^n$ and $\displaystyle c^n$. We know that for any values greater than 2, $\displaystyle a^n + b^n ≠ c^n$.
No, we don't! As I said before, we only know that there are no integer values of a, b, and c satisfying that. There certainly are real number values.

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However, is there a general formula that would input $\displaystyle a^n + b^n$ and return/output $\displaystyle c^n$?
Of course there is: take a to the nth power, take b to the nth power and add! That return c^n. And, as Maschke said, you can get c by taking the nth root of that. If a= b= 1, then c is the nth root of 2.
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October 14th, 2017, 05:09 AM   #7
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Quote:
you can get c by taking the nth root of that.
Possibly it is worth making the distinction between the nth root function, which is single valued by definition and the nth root itself which may be one of several values.

So the sentence might be more accurately phrased "bytaking an nth root....."
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October 15th, 2017, 12:14 PM   #8
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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.
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October 15th, 2017, 12:17 PM   #9
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Beg to differ w/Country Boy and studiot.

OP specified triangles. We're talking about positive numbers. The n-th root is unambiguous.

Last edited by Maschke; October 15th, 2017 at 12:22 PM.
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