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I have a question regarding the following problem.

Let $A,B,C$ be the three independently selected, uniformly distributed points on the unit sphere $S^3$ in $\Bbb R^4$. What's the probability that the unique chordal triangle $ABC$ is acute?

I just started analysis, and I have little knowledge about metric space.(I just learned the definition of the metric space.)

I'm wondering if it is okay to change "the unit sphere $S^3$ in $\Bbb R^4$" to "the unit sphere $S^3$ in $\Bbb R^3$".

Thanks for your help.

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    It is not: $S^3$ is not a subset of $\Bbb R^3$.2012-08-10
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    I thought S^3 is x^2+y^2+z^2=1 and R^3 is just xyz?2012-08-10
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    $S^3=\{\langle w,x,y,z\rangle\in\Bbb R^4:w^2+x^2+y^2+z^2=1\}$. Similarly, $S^2=\{\langle x,y,z\rangle\in\Bbb R^3:x^2+y^2+z^2=1\}$, and $S^1=\{\langle x,y\rangle\in\Bbb R^2:x^2+y^2=1\}$.2012-08-10
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    By the way, I don't know how to use LaTeX. For example, my comment is just "s^3" unlike yours. What's the difference?2012-08-10
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    I got it. Thank you very much.2012-08-10
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    Put your cursor on a formula and right-click. Select *Show Math As* and then *TeX Commands*; you’ll get a pop-up window showing the original $\LaTeX$ code.2012-08-10
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    Note that the dollar signs aren't included in that window, though; you need to enclose any $\TeX$ code you use in dollar signs; single dollar signs for inline formulas, double dollar signs for displayed equations.2012-08-10
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    @Brian M. Scott Thank you for your reply. Okay I'll try it.2012-08-10
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    @joriki Thank you for your comment. Let me try that. $S^2$ $cosx$2012-08-10
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    Ohh it worked thanks2012-08-10
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    I don't think your question doesn't really have anything to do with any metrics. Except perhaps for the one in the definition of unit sphere, but that does not really warrant tagging it as metric-spaces...2012-08-10
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    @BrianM.Scott: and $S^0=\{x\in \mathbf R\vert x^2=1\}$, if you want to be that thorough. ;)2012-08-10
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    @tomasz It seems like I was confused because I did not know the definition of unit sphere.2012-08-11

1 Answers 1

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I'll assume that by "the unique chordal triangle $ABC$" you mean simply the triangle $ABC$.

A triangle can have at most one obtuse angle. Thus the probability for it to have an obtuse angle is three times the probability of a given one of the three angles being obtuse. Thus we just have to calculate that probability.

Imagine $A$ and $B$ having been chosen, and place a pole at the midpoint between them. Then the angle $ACB$ will be obtuse if and only if $C$ is chosen closer to the pole than $A$ and $B$. This yields a double integration for the desired probability:

$$ \begin{align} P(\angle ACB\text{ obtuse}) &= \frac1{S_{d+1}^2}\int_0^\pi\mathrm d\theta\,S_d\,\sin^d\theta\int_0^{\theta/2}\mathrm d\phi\,S_d\,\sin^d\phi \\ &=\left(\frac{S_d}{S_{d+1}}\right)^2\int_0^\pi\mathrm d\theta\,\sin^d\theta\int_0^{\theta/2}\mathrm d\phi\,\sin^d\phi\;, \end{align} $$

where $d$ is one less than the dimension of the sphere (in your case $d=2$), $\theta$ is the angular distance between $A$ and $B$, $S_d$ is the surface area of the unit $d$-sphere, $S_d\,\sin^d\theta$ is the angular density of points $B$ at angular distance $\theta$ from $A$, $\theta/2$ is the angular distance from the pole to $A$ and $B$, $\phi$ is the angular distance from the pole to $C$ and $S_d\,\sin^d\phi$ is the angular density of points $C$ at angular distance $\phi$ from the pole.

To warm up, we can recover the known result $1/4$ for $d=0$ (a circle):

$$ \left(\frac{S_0}{S_1}\right)^2\int_0^\pi\mathrm d\theta\int_0^{\theta/2}\mathrm d\phi=\left(\frac{2}{2\pi}\right)^2\int_0^\pi\mathrm d\theta\,\frac{\theta}2=\frac14\;. $$

Next, with $d=1$, the result for a sphere:

$$ \left(\frac{S_1}{S_2}\right)^2\int_0^\pi\mathrm d\theta\,\sin\theta\int_0^{\theta/2}\mathrm d\phi\,\sin\phi = \left(\frac{2\pi}{4\pi}\right)^2\int_0^\pi\mathrm d\theta\,\sin\theta\left(1-\cos\frac\theta2\right)=\frac16\;. $$

And now, with only slightly more involved integrals, your case $d=2$:

$$ \begin{align} \left(\frac{S_2}{S_3}\right)^2\int_0^\pi\mathrm d\theta\,\sin^2\theta\int_0^{\theta/2}\mathrm d\phi\,\sin^2\phi &= \left(\frac{4\pi}{2\pi^2}\right)^2\int_0^\pi\mathrm d\theta\,\sin^2\theta\left[\frac12(\phi-\sin\phi\cos\phi)\right]_0^{\theta/2} \\ &= \frac2{\pi^2}\int_0^\pi\mathrm d\theta\,\sin^2\theta\left(\frac\theta2-\sin\frac\theta2\cos\frac\theta2\right) \\ &= \frac14-\frac4{3\pi^2}\;. \end{align} $$

This is the probability for a given angle to be obtuse; the probability for any angle to be obtuse is three times that, and so the probability for the triangle to be acute is

$$ 1-3\left(\frac14-\frac4{3\pi^2}\right)=\frac14+\frac4{\pi^2}\approx0.6553\;. $$

(I checked these results with numerical simulations.)