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 Subject: Tetrahedron solid angle Category: Science > Math Asked by: racecar-ga List Price: \$2.00 Posted: 11 Feb 2003 17:48 PST Expires: 13 Mar 2003 17:48 PST Question ID: 160247
 ```What is the solid angle subtended by one vertex of a regular tetrahedron? Ideal answer: a number and an explanation.``` Request for Question Clarification by mathtalk-ga on 14 Feb 2003 18:27 PST ```Hi, racecar-ga: The numeric answer provided by foodini-ga's comment is correct. I was wondering if you still wanted more in the way of explanation about this number. regards, mathtalk-ga``` Clarification of Question by racecar-ga on 14 Feb 2003 21:22 PST ```Foodini's answer is based on the formula for the area of a triangle on a unit sphere: A = a + b + c - pi where a, b, and c are the angles of the triangle. I would be thrilled to know the derivation of this formula.```
 Subject: Re: Tetrahedron solid angle Answered By: mathtalk-ga on 15 Feb 2003 10:53 PST Rated:
 ```Hi, racecar-ga: The formula for area of a spherical triangle is known as Girard's theorem: area = (a + b + c - pi)*R^2 if the radius of the sphere is R. In your application the radius of the sphere is 1, but it may be helpful to leave the factor R^2 intact for the time being so the units of area are apparent in the formulas. (We will measure angles in radians throughout.) John C. Polking of Rice University maintains this nice Web site: [The Geometry of the Sphere] http://math.rice.edu/~pcmi/sphere/ where a proof of this and other "sphere facts" are illustrated by interactive Java applets for clarity. First the area of the whole sphere A = 4pi*R^2 is credited to Archimedes: [An ancient extra-geometric proof] http://www.cut-the-knot.com/pythagoras/Archimedes.shtml See Note 1 below for further comments on this. From the area of the whole sphere one deduces the area of a "lune" (region bounded by segments of two great circles). Because the two points at which two great circles intersect are "antipodal" (at opposite poles), one can visualize dividing the entire sphere into some whole number q of parts, for which the area L of the lune is related to the angle 2pi/q between the great circles in this manner: L = (4pi*R^2)/q = 2(2pi/q)*R^2 By combining several such "integral subdivisions" of the whole sphere, one has for any rational p/q the more general case of a lune with angle 2pi(p/q): L = (4pi*R^2)*(p/q) = 2(2pi(p/q))*R^2 One then argues by continuous dependence of area of the lune on this angle that in general (since rational numbers can approximate any real number with arbitrary accuracy): L = 2a*R^2 when a is the angle between the great circles at either "corner" of the lune. For reference that argument is given here: [The area of a lune] http://math.rice.edu/~pcmi/sphere/gos3.html#1 With these two facts (area of the sphere, area of a lune) we have the ingredients for proving Girard's Theorem: [The area of a spherical triangle] http://math.rice.edu/~pcmi/sphere/gos4.html#1 Let T be a spherical triangle with angles a,b,c at its three corners. Extend the great circles which form the sides of T around the sphere, and these meet in a congruent spherical triangle T' on the opposite side of the sphere. Indeed these great circles bound three triangles which share a side with T and meet T' at a corner and three other triangles which share a side with T' and meet T at a corner. What we will describe next may be difficult to visualize without illustration, so by all means take a look (with a Java enabled browser) at the nice 3D diagrams presented at the Web site above. Just as T may be said to be antipodal to T', each of the first three triangles is paired with one of the last three. Together T,T' and such corresponding pair of other triangles form two lunes, each with an angle a, b, or c taken from the respective corner of T. Altogther then there are six lunes, and we can add up their areas in two different ways. On one hand we can simply add the areas of three pairs of lunes individually: 2 * [2a*R^2 + 2b*R^2 + 2c*R^2] = 4(a + b + c)*R^2 On the other hand the six lunes collectively cover each part of the sphere once except for T,T' which are covered three times apiece. So the combined areas of these six lunes is also the area of the sphere plus two (extra) copies of T,T': 4pi*R^2 + 2*(area of T) + 2*(area of T') Now T,T' have equal areas, so by equating the above expressions and solving for the area of T: 4*(area of T) = 4(a + b + c)*R^2 - 4pi*R^2 area of T = (a + b + c - pi)*R^2 Thus Girard's Theorem is proven. See Note 2 below about who Girard was. For the sake of completeness we recap the calculation initially proposed by foodini in answer to your question. Consider a regular tetrahedron with one vertex at the center of a unit sphere and the other three vertices on the sphere's surface. The three angles of the spherical triangle formed by those latter three vertices are equal, and we can compute them as dihedral angles (angles between two sides of the tetrahedron). Drop a perpendicular from one vertex of the tetrahedron to the median of the opposite equilateral triangular face. This "tetrahedral" altitude forms a right triangle with a shorter leg that is 1/3 of the altitude of that triangular face and a hypotenuse (on an adjacent face) that is an entire "triangular" altitude. Hence, as foodini-ga claims, the angle between those two triangular faces has cosine of 1/3. Since now R = 1, we have the solid angle as: area of spherical triangle = 3 arccos(1/3) - pi = 0.551285598432530807942144151464459... steradians regards, mathtalk Note 1: [Archimedes on Circles and Spheres (scroll to bottom)] http://www.maths.uwa.edu.au/~schultz/3M3/L7Archimedes1.html Archimedes' "patented" Method of Exhaustion is more or less equivalent to our modern day integral calculus, i.e. sum up a series of frustums (banded sections) of cones to give the area of a hemisphere with ever increasing accuracy. Archimedes also pointed out the relationship of sphere volume V to sphere surface area A: V = (1/3)*R*A which today we'd likely present in terms of differentiating V with respect to R, giving A. A = dV/dR = 3V/R Note 2: Girard's Theorem is named after Albert Girard (1595-1632), a French born mathematician who moved to the Netherlands for religious refuge: [Mathematicians: Albert Girard] http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Girard_Albert.html He was first to use the now familiar abbreviations sin, cos, tan for elementary trigonometric functions. Note 3: A more general formula for the solid angle subtended by a triangle is recounted here by A. van Oosterom: [Re: Solid angle subtended by rectangle] http://mathforum.org/epigone/geometry-puzzles/crurdrauku/jpxw4tv0za99@legacy Let U,V,W be three vectors emanating from the origin, with respective lengths u,v,w. The (signed) area corresponding to the solid angle formed at this corner is: 2 arctan( det(M)/(uvw + u V.W + v U.W + w U.V) ) where M is the matrix formed by stacking U,V,W as rows and the period "." is used to denote the dot product between vectors. Observe that the denominator here is not affected by the ordering of the three vectors, but the sign of the numerator alternates with transpositions of any two vectors. To duplicate the earlier numerical result one takes unit vectors: U = (1,0,0) V = (1/2,sqrt(3)/2,0) W = (1/2,sqrt(3)/6,sqrt(6)/3) det(M) = sqrt(2)/2 uvw + u V.W + v U.W + w U.V = 1 + 1/2 + 1/2 + 1/2 2 arctan((sqrt(2)/2)/(5/2)) = 0.551285598432530807942144151464459... as previously found.```
 racecar-ga rated this answer: and gave an additional tip of: \$3.00 `Fabulous answer. Thank you very much.`

 ```I think the page http://www.cs.williams.edu/~98bcc/tiling/danztable.html might be of use to you. The solid angle subtended by the corner of a tetrahedron should be the angles the planes form with each other, times three, minus pi. That's it. How do I collect my two bucks. ;)```
 ```Doh. Forgot: The dihedral angles of the planes of a tetrahedron are arcos(1/3), making the solid angle of the corner of a tetrahedron 3*(arcos(1/3)) steradians, or roughly .55128 steradians.```
 ```Thanks foodini, this seems right. Now I just wish I knew that 'simple and elegant proof' referred to in the link you provided.```
 ```Hi, racecar-ga: This is off-topic in relation to this question, but on the now expired: http://answers.google.com/answers/main?cmd=threadview&id=240017 you wondered if I would "bite". I refrained from posting because I believe lynnspry-ga's link gives you the best possible answer. For any possible number of hat colors there is a deterministic strategy in which at most one prisoner (the first) guesses incorrectly. [Presumably the first prisoner has no information about his/her own hat color, so the chance of any guess being correct is 1/n.] Let's walk through the case of three hats in detail, because I fail to see anything "cheesy" about the solution. Let the three colors be numerically coded mod 3: Red = 0 Blue = 1 Green = 2 (or equivalently, -1) The first prisoner looks out over the sea of hats ahead and sums the visible hat colors mod 3. In this case it amounts to taking: (# of Blue hats) - (# of Green hats) mod 3 The first prisoner announces as his or her guess the color (Red, Green, or Blue) that corresponds to the numeric value (0,1,2 mod 3) obtained in this way. All the prisoners need to hear and remember the first prisoner's guess, as well as the subsequent "guesses" of any other prisoners who precede them. The cumulative information allows each prisoner in turn to determine the color of their own hat accurately. Here's how. The second prisoner mentally calculates the same quantity above, but looking at all the hats that the first prisoner saw _except_ his or her own hat. Now the two sets of hats involved differ only by the second prisoner's hat, so there are three and only three possibilities: 0) The counts are equal. This means the second prisoner's hat is Red (since the difference was zero). The second prisoner says Red (which is correct), and all the other prisoners mentally "update" the "count" to stay the same. 1) The second prisoner's count is 1 more than the first prisoner's count (ie. taken mod 3, the count goes from 0 --> 1, or from 1 --> 2, or from 2 --> 0). This means that the "missing" hat color (the one worn by the second prisoner) must be Green (of value 2 = -1 mod 3, since the count increased by 1). The second prisoner says Green (which is correct), and all the remaining prisoners mentally update the count to be increased by 1 mod 3. 2) The second prisoner's count is 1 less than the first prisoner's count (ie. taken mod 3, the count goes from 0 --> 2, or from 1 --> 0, or from 2 --> 1). This means that the "missing" hat color (the one worn by the second prisoner) must be Blue (of value 1 mod 3, since the count decreased by 1). The second prisoner says Blue (which is correct), and all the remaining prisoners mentally update the count to be decreased by 1 mod 3. The approach then applies recursively to all prisoners down the line. For the sake of completeness, let's consider the final prisoner, who sees no hats in front and therefore counts zero in all circumstances. By keeping a running count of: (# of Blue hats) - (# of Green hats) mod 3 up to this point, the final prisoner has enough information to deduce the final hat color. Namely if that count (as deduced from the responses of the next to last prisoner and all those who preceded) were 0, it could only mean the final hat color was Red. Alternatively, that count would be +1 mod 3 only if the final hat was Blue, and +2 mod 3 only if the final hat was Green. It is striking to me that the same approach is optimal (for the prisoner's group mortality) regardless of the number of hat colors. The first prisoner has no information (presumably) about his or her own hat color, so one guess is as good as any other (assuming the hat colors are independently chosen, so that seeing all other hats provides no clue about one's own hat color). The first prisoner therefore "sets free" all the other prisoners (who will assuredly act in their own best interest) without making any "extra" sacrifice (just a commitment to follow the formula, which has as much chance of producing a correct guess as any other method, namely 1/n). regards, mathtalk```