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Description: consider a regular tetrahedron (with height 1), construct a sphere centering at one of the tetrahedron's vertex, with radius 1 also. Then what's the surface area of the sphere's portion that gets cut out by the tetrahedron? Following are links to some illustrations I made using Mathematica.

$\hskip 1.3in$ a side view $$\text{a side view}$$

$\hskip 1.3in$ a bottom view: you are viewing the unknown area directly $$\text{a bottom view: you are viewing the unknown area directly}$$

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    $\begingroup$ I studied this problem years ago, it took me quite a long time to realize what aspects of information I should look for. The key word is SPHERICAL TRIGONOMETRY, I'm sure you'll find useful information from Google. :) $\endgroup$ – A. Chu Jun 5 '13 at 9:05
  • $\begingroup$ @ᴊᴀsᴏɴ Honestly, this problem has been in my mind for years. I thought about spherical trigonometry but still had no idea how to start. I am going to study the following detailed answers and get this mystery solved :-) $\endgroup$ – Taozi Jun 5 '13 at 15:04
  • $\begingroup$ @achillehui Thank you, and thanks to Mathematica~ $\endgroup$ – Taozi Jun 5 '13 at 15:06
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One useful tool in getting this kind of information is the solid angle of an object. Simply multiplying such a solid angle by $r^2$ will give you the surface area cut out from a sphere of radius $r$ centered around the apex of the angle. Since your sphere has radius $1$, you're in fact simply asking for the solid angle at the top of a tetrahedron. Wikipedia has a section on its computation. That section is for general, i.e. not neccessarily regular tetrahedra. You could apply those computations to a regular tetrahedron. Or you have a look at the article on (the regular) tetrahedron. There you find the solid angle at a vertex to be

$$\Omega = \arccos\frac{23}{27} \approx 0.55129\,\text{sr}$$

Since your radius is $1$, you can ignore the unit of sr and interpret this number directly as the surface area you asked about. Note that the size of the tetrahedron is irrelevant in all of this, as long as one of its vertices is centered at the center of the sphere and the opposite side lies completely outside the sphere.

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  • $\begingroup$ The first link gives an answer the same as Christian Blatter's solution above. The second link has omega = arccos(23/27) which seems to be obtained by some other methods, or just a pure algebraic transformation of the former. Thank you for the links. $\endgroup$ – Taozi Jun 7 '13 at 8:34
  • $\begingroup$ @Taozi, just addition formula for cosine.$$\cos\Omega = \cos(3\theta-\pi) = -\cos(3\theta) = 3\cos\theta - 4\cos\theta^3 = 3(\frac13)-4(\frac13)^3 = \frac{23}{27}$$ $\endgroup$ – achille hui Jun 7 '13 at 9:19
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The tetrahedron $T$ having height $1$ means that one face of $T$ just touches the sphere. Therefore the set $T\cap S^2$ doesn't have a hole and is an equilateral spherical triangle $\Delta$ with angles $\theta$. By a standard formula of spherical trigonometry the area of $\Delta$ is the sum of its angles minus $\pi$, so it is given by $$\omega(\Delta)=3\theta- \pi\ .$$ Therefore it remains to determine this angle $\theta$. Since $S^2$ intersects the three edges of $T$ emanating from $0$ orthogonally it follows that $\theta$ is equal to the angle between any two adjacent faces of $T$. Let $A$ and $B$ be two vertices of $T$ and $M$ the midpoint of the opposite edge. Then $\theta$ is the angle at the tip $M$ of the isosceles triangle $ABM$, and it follows that $$\sin{\theta\over2}={s/2\over s\ \sqrt{3}/2}\ ,$$ or $\theta=2\arcsin{1\over\sqrt{3}}$. It follows that $$\omega(\Delta)=6\arcsin{1\over\sqrt{3}}-\pi\doteq0.551286\ .$$

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  • $\begingroup$ Thanks! I reviewed some spherical geometry and now understand your answer, indeed, without some knowledge of the geometry the result appears like black magic :-) $\endgroup$ – Taozi Jun 7 '13 at 8:27

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