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like in pythagorean theorem angle comes 90 degree for the expression $a^2 + b^2 = c^2$, however I know that no integer solution is possible.

Old John
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Rahul
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4 Answers4

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There is no single angle corresponding to the relationship $a^3+b^3=c^3$.

Suppose that a triangle has sides of lengths $a,b$, and $c$ such that $a^3+b^3=c^3$. We know from the law of cosines that if $\theta$ is the angle opposite the side of length $c$, then $c^2=a^2+b^2-2ab\cos\theta$, so

$$\cos\theta=\frac{a^2+b^2-c^2}{2ab}\;.$$

Now let’s look at just a few examples. If $a=b=1$, then $c=\sqrt[3]2$, and $$\cos\theta=\frac{2-2^{2/3}}2\approx0.20630\;.$$

If $a=1$ and $b=2$, then $c=\sqrt[3]9$, and $$\cos\theta=\frac{5-9^{2/3}}4\approx0.16831\;.$$

If $a=1$ and $b=3$, then $c=\sqrt[3]{28}$, and $$\cos\theta=\frac{10-28^{2/3}}6\approx0.12985\;.$$

As you can see, these values of $\cos\theta$ are all different, so the angles themselves are also different.

Brian M. Scott
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And here's a plot of the angle:

1

The minimum angle seems to be at $a = b$, where

$$\cos\theta = 1 - \frac{1}{2^{1/3}}$$

wxffles
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1

You do get slightly different answers for the Pythagorean Theorem on the surface of a sphere of radius $1,$ or the hyperbolic plane of curvature $-1.$ On the sphere, a right triangle with geodesic lengths of legs $a,b$ and hypotenuse $c$ obeys $$ \cos c = \cos a \; \cos b, $$ while in the hyperbolic plane with curvature $-1$ it becomes $$ \cosh c = \cosh a \; \cosh b. $$ In both cases, if you write out the functions as power series in $a,b,c,$ you see that the limit as $a,b,c$ all shrink to nearly $0$ is the traditional Pythagorean Theorem.

See Wikipedia Section Link

Will Jagy
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  • Wookieepedia: The article you requested does not exist. –  Aug 19 '12 at 21:40
  • @RahulNarain, I see what you mean. For some reason mine works, but I can understand that it is not really a Star Wars kind of thing. Life is so complicated. – Will Jagy Aug 19 '12 at 21:45
  • I am very confused about what "WOOKIE" means. Presumably you're not referring to Star Wars? – Potato Aug 19 '12 at 22:12
  • @Potato, it is just a link to a relevant section in Wikipedia. I have paid little attention to these changes in the web over the years, and to me Wiki might as well be random syllables, to be fiddled with for my amusement. However, if it is actually going to confuse people, I suppose I can write this is a more standard manner. On my screen, any link appears among text in a separate blue color. – Will Jagy Aug 19 '12 at 22:28
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    @WillJagy I just thought it was an acronym and was wondering what it stood for. – Potato Aug 19 '12 at 22:53
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Expanding on Brian M. Scott's answer, since $\cos\theta=\frac{a^2+b^2-c^2}{2ab}$ and $c^3 = a^3+b^3$, $\cos\theta=\frac{a^2+b^2-(a^3+b^3)^{2/3}}{2ab} = \frac{1+(b/a)^2-(1+(b/a)^3)^{2/3}}{2} =\frac{1+r^2-(1+r^3)^{2/3}}{2} $ where $r = b/a$.

The derivative of the numerator is $2r-(2/3)(3r^2)(1+r^3)^{-1/3} = 2r - 2r^2(1+r^3)^{-1/3}$ which is never 0 (else $(1+r^3)^{1/3} = r$) so is always positive (since its value at 1 is $2-2/2^{1/3} > 0$).

For large r, $(1+r^3)^{2/3} = r^2 (1+r^{-3})^{2/3} \approx r^2(1 + (2/3)r^{-3}) = r^2 + 2/(3r) $ so $\cos\theta \approx \frac{1+r^2 - (r^2 + 2/(3r))}{2} = \frac{1-2/(3r)}{2} $ which tends to 1/2 for large $r$.

This may have an error, since I would expect it to go to 1, but I have to go now, so this this is it.

marty cohen
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