Assignment 10
Problems
Platonic Solids
In these problems we will investigate regular polygons on the sphere. Recall we call a polygon regular if it has rotational symmetries about its center: in particular this implies that all its sides are the same length, and all its angles have the same measure (since isometries preserve both lengths and angles).
In the Euclidean plane, we know that regular polygons of all side numbers \(\geq 3\) exist (these are how Archimedes approximated the circle, after all!), but their angles are strictly determined by their number of sides. We proved in a previous homework that the angle sum of an \(n\)-gon is \((n-2)\pi\), and if all the angles of a regular \(n\) gon are equal, each angle must measure \(\theta_n = \frac{n-2}{n}\pi\).
This puts a strong restriction on which regular polygons can be used to tile the plane. To tile the plane, a necessary (but not sufficient) condition is that we need to be able to fit \(k\) copies of each polygon around a vertex, without any gaps or overalps. This tells us that the angles of a polygon that can tile must be \(\theta =\tfrac{2\pi}{k}\).
Thus, to figure out which polygons even have a chance of tiling the Euclidean plane, we want to know for which \(n\) (the number of sides) there the angle \(\theta_n\) is actually \(2\pi\) over an integer. We can start listing:
\[\theta_3=\frac{3-2}{3}{\pi}=\frac{\pi}{3}=\frac{2\pi}{6}\] \[\theta_4 = \frac{4-2}{4}\pi=\frac{\pi}{2}=\frac{2\pi}{4}\] \[\theta_5=\frac{5-2}{5}\pi=\frac{3\pi}{5}\] \[\theta_6=\frac{6-2}{6}{\pi}=\frac{2\pi}{3}\] \[\theta_7=\frac{7-2}{7}{\pi}=\frac{5\pi}{7}\]
Thus, we see that its possible to fit six triangles around a vertex, four squares around a vertex and three hexagons around a vertex, but as the angles \(\theta_5\) and \(\theta_7\) aren’t even divisions of \(2\pi\), there’s no nice way to fit pentagons or 7-gons around a vertex, and thus no hope of using them to tile the plane.
This is the start to the classification of regular tilings of the plane, where by what we see from the angle measures, its possible for triangles, squares and hexagons, but impossible for all other shapes!
Our goal here is to investigate what changes on the sphere.
Exercise 1 (Spherical Pentagons)
Find a relationship between the area \(A\) of a spherical regular pentagon and its angle measure \(\alpha\). Hint: divide the spherical pentagon into five triangles
Show that there exists a spherical pentagon whose angle evenly divides \(2\pi\): how many of these spherical pentagons fit around a single vertex?
What is the area of such a spherical pentagon? How many of these pentagons does it take to cover the entire sphere?
For the pentagon, there was only one possibility, as the restriction that the angle at a vertex be \(2\pi/k\) is so restrictive. Howver, for triangles, there are three possibilites!
Exercise 2 (Spherical Triangles) There are three different equilateral triangles that can be used to tile the sphere. Find them! For each triangle:
- How many fit around each vertex?
- How many are needed to cover the sphere?
- What platonic solid does this correspond to?
The Pythagorean Theorem
The fundamental formula in Euclidean trigonometry is the Pythagorean theorem which allows us to measure the length of the hypotenuse of a right triangle in terms of the other side lengths.
The goal of this exercise is to derive the spherical counterpart to this:
Theorem 1 (Spherical Pythagorean Theorem) Given a right triangle on \(\SS^2\) with side lengths \(a,b\) and hypotenuse \(c\), these three lengths satisfy the equation \[\cos(c)=\cos(a)\cos(b)\]
Exercise 3 (Deriving The Pythagorean Theorem) Prove that the formula given above really does hold for the legs and hypotenuse of a right triangle on \(\SS^2\), using the distance formula that we’ve already calculated:
\[\cos\dist(p,q)=p\cdot q\]
Hint: move your triangle so the right angle is at the north pole, and the legs are along the great circles on the \(xz\) and \(yz\) plane. Now you can write down exactly what the other two vertices are since you know they are distance \(a\) and \(b\) along these geodesics from \(N\), and these geodesics are unit circles in the \(xz\) and \(yz\) planes!*
On a sphere of radius \(R\), a similar formula exists: here to be able to use arguments involving angles we need to divide all the distances by the sphere’s radius, but afterwards an argument analogous to the above exercise yields
\[\cos\left(\frac cR\right)=\cos\left(\frac aR\right)\cos\left(\frac bR\right)\]
Its often more useful to rewrite this result in terms of the curvature \(\kappa=1/R^2\)
Theorem 2 (Pythagorean Theroem of Curvature \(\kappa\)) On the sphere of curvature \(\kappa\), the two legs \(a,b\) and the hypotenuse \(c\) of a right triangle satisfy \[\cos\left(c\sqrt{\kappa}\right)=\cos\left(a\sqrt{\kappa}\right)\cos\left(b\sqrt{\kappa}\right)\]
As a sphere gets larger and larger in radius, it better approximates the Euclidean plane. We might even want to say something like in the limit \(R\to\infty\) (so, \(\kappa\to 0\)) the spherical geometry becomes euclidean. But how could we make such a statement precise? One way is to study what happens to the theorems of spherical geometry as \(\kappa\to 0\); and show that they become their Euclidean counterparts. The exercise below is our first encounter with this big idea:
Exercise 4 (Euclidean Geometry as the Limit of Shrinking Curvature) Consider a triangle with side lengths \(a,b,c\) in spherical geometry of curvature \(\kappa\). As \(\kappa\to 0\), the arguments of the cosines in the Pythagorean theorem become very small numbers, so it makes sense to approximate approximate these with the first terms of their Taylor series.
Compute the Taylor series of both sides of \[\cos\left(c\sqrt{\kappa}\right)=\cos\left(a\sqrt{\kappa}\right)\cos\left(b\sqrt{\kappa}\right)\]
in the limit \(\kappa\to 0\), we can ignore all but the first nontrivial terms. Show here that only keeping up to the quadratic terms on each side recovers the Euclidean Pythagorean theroem, \(c^2=a^2+b^2\).
Trigonometry
Following the derivation of the spherical pythagorean theorem, we might next hope to discover relationships between the sides of a spherical right triangle and its angle measures. And, indeed we can!
The corresponding laws of spherical trigonometry are as follows:
Theorem 3 (Spherical Trigonometric Relations) For a right triangle with angles \(\alpha,\beta\), corresponding opposite sides \(a,b\) and hypotenuse \(c\) the following relations hold:
\[\sin \alpha = \frac{\sin a}{\sin c}\hspace{1cm}\sin\beta =\frac{\sin b}{\sin c}\]
\[\cos\alpha = \frac{\tan b}{\tan c}\hspace{1cm}\cos\beta=\frac{\tan a}{\tan c}\]
You will not be responsible for the derivation of these formulas, nor for remembering them: if you ever need them they will be given to you!
One of the most biggest differences between spherical trigonometry and its Euclidean counterpart is that its possible to derive formulas for the length of a triangles’ sides in terms of only the angle information! This is impossible in Euclidean space because of the existence of similarities: there are plenty of pairs of triangles that have all the same angles but wildly different side lengths! No so in the geometry of the sphere.
Exercise 5 Using the trigonometric identities in Theorem Spherical Trigonometric Relations together with the spherical pythagorean theorem Theorem Spherical Pythagorean Theorem, show that the side length \(a\) of a right triangle can be computed knowing only the opposite angle \(\alpha\) and the adjacent angle \(\beta\) as
\[\cos a = \frac{\cos\alpha}{\sin\beta}\]
Hint: start with the formula for \(\cos\alpha\). Write out the tangents in terms of sines and cosines, then apply the pythagorean theorem to expand a term. Finally, use the definition of \(\sin\beta\) to regroup some terms.
Formulas such as this are incredibly useful for calculating the side lengths of polygons, by dividing them into triangles and using facts that are known about their angles.
Exercise 6 (Spherical Trigonometry) Use spherical trigonometry to figure out the side lengths of the pentagon you discovered in the first exercise.
Hint: can you further divide the five triangles you used before, into ten right triangles inside the pentagon?
Solutions
Platonic Solids
Spherical Pentagons
Given a regular spherical pentagon with all angles equal to \(\alpha\), we may divide the pentagon into five triangles by placing a point in the center and connecting it to all five vertices. The resulting triangles have a pair of angles of measure \(\alpha/2\) on the outside, and a single angle at the interior vertex of \(2\pi/5\).
PICTURE
Thus, the area of such a triangle is (\(2\pi/5 +\alpha/2 +\alpha /2)-pi = \alpha -\frac{3\pi}{5}\), and the area of the entire pentagon is five times this:
$\(\area = 5\alpha -3\pi\)
Because we want \(n\) pentagons to fit around a vertex to make a tiling (where \(n\geq 3\), as we cant have just one or two polygons meeting at a single point), we can set \(\alpha=2\pi/n\) and then we wish to figure our for which values of \(n\) such a pentagon exists. For a pentagon to exist it must have positive area, so we can use the formula above to help us out: we want
\[5\frac{2\pi}{n}-3\pi >0\]
Getting a common denominator this imples \((10\pi-3\pi n)/n>0\) which is equivalent to \(10>3n\). From here, we see directly that ony \(n=3\) works, as if \(n\geq 4\) then \(3n/geq 12\) is certainly not less than \(10\). In the working case \(n=3\), the area of the pentagon is \[5\frac{2\pi}{3}-3\pi = \frac{\pi}{3}\]
And since the total area of the unit sphere is \(4\pi\), this means we can fit 12 such pentagons on the sphere. This is the dodecahedron: twelve pentagons meeting three to a vertex!
Triangle Tilings
The solution to this problem is very analgous to the above, but instead of working with pentagons we are looking directly at triangles. We find there are three solutions, one with four triangles (the tetrahedron) one with eight triangles (the octahedron) and one with twenty triangles (the icosahedron).
Pythagorean Theorem
Derivation
Let \(T\) be a right triangle on the sphere with legs of length \(a,b\) and a hypotenuse of length \(c\). By homogenity of the sphere, we can use an isometry to move the vertex of \(T\) where the legs meet to the north pole \(N\). Then, by isotropy of the sphere we can rotate the sphere around \(N\) so that the first leg lies on the great circle in the \(xz\) plane. Becuase the triangle is a right triangle, the other leg must then lie on the great circle in the \(yz\) plane.
PICTURE
Now, we can write down in coordiantes the location of these other two vertices. Since the first lies at distace \(a\) along the great circle in the \(xz\) plane, its coordinates must be \[p=(\sin a,0,\cos a)\] Similarly, since the second lies at distance \(b\) from \(N\) along the great circle in the \(yz\) plane, its coordinates must be \[q=(0,\sin b,\cos b)\]
We derived in class that the distance function between two points \(p,q\) on the sphere satisfies \(\cos(\mathrm{dist})=p\cdot q\). Applying that here, since we know the distance is \(c\) (its the hypotenuse), we have
\[\cos c = (\sin a,0,\cos a)\cdot (0,\sin b,\cos b)=\cos a\cos b\]
The Limit
Here our goal is to Taylor expand the pythagorean theorem in curvature \(\kappa\), and then simplify so that we can take the limit as \(\kappa\to 0\). We will use the function \(\cos(x\sqrt{\kappa})\) multiple times, so we record its series expansion here:
\[\cos(x\sqrt{\kappa})=1-\frac{k x^2}{2}+\frac{k^2x^4}{4!}-\cdots\]
Thus, the pythagorean theorem expands to
\[1-\frac{k c^2}{2}+\frac{k^2c^4}{4!}-\cdots=\left(1-\frac{k a^2}{2}+\frac{k^2a^4}{4!}-\cdots\right)\left(1-\frac{k b^2}{2}+\frac{k^2b^4}{4!}-\cdots\right)\] \[=1-\frac{ka^2}{2}-\frac{kb^2}{2}+\frac{k^2a^2b^2}{2\cdot 2}+\frac{k^2a^4}{4!}+\frac{k^2b^4}{4!}-\cdots\]
We can simplify this a bit by seeing each side has a 1, so we can subtract it; and then multiply everything by \(-1\) to make the beginning terms positive (just to be easier to work with).
\[\frac{kc^2}{2}-\frac{k^2c^4}{4!}+\cdots = \frac{ka^2}{2}+\frac{kb^2}{2}-\frac{k^2a^2b^2}{2\cdot 2}-\frac{k^2a^4}{4!}-\frac{k^2b^4}{4!}+\cdots\]
Now, all these terms have at least one factor of \(k\) in them, so we can cancel this from both sides:
\[\frac{c^2}{2}-\frac{kc^4}{4!}+\cdots = \frac{a^2}{2}+\frac{b^2}{2}-\frac{ka^2b^2}{2\cdot 2}-\frac{ka^4}{4!}-\frac{kb^4}{4!}+\cdots\]
Now we may take the limit as k$ without any issue. This just makes all the terms with a \(k\) in them disappear, leaving us with
\[\frac{c^2}{2}=\frac{a^2+b^2}{2}\]
Multiplying by \(2\) yeilds what we expect, the Euclidean pythagorean theorem \[c^2=a^2+b^2\]
Trigonometry
Lengths in Terms of Angles
We know that \(\cos\alpha = \frac{\tan b}{\tan c}\), and expanding these tangents gives
\[\cos\alpha = \frac{\sin b/\cos b}{\sin c/\cos c}=\frac{\sin b\cos c}{\cos b\sin c}\]
Using the pythagorean theorem \(\cos c=\cos a\cos b\) lets us re-express this as
\[\cos \alpha = \frac{\sin b\cos a\cos b}{\cos b\sin c}=\frac{\sin b\cos a}{\sin c}\]
Then recalling that in spherical trigonometry we have \(\sin \beta = \sin b/\sin c\) this simplifies to what we wanted:
\[\cos\alpha = \sin \beta\cos a\,\,\implies\,\,\cos a=\frac{\cos\alpha}{\sin\beta}\]
This is a rather striking formula, that has no analog in Euclidean geometry: it tells us that its poasible to determine the side lengths of a triangle if all we know is the angles!
The Side Length of a Pentagon
We can use this newfound formula to determine the side lengths of the pentagons making up our spherical dodecahedron. Before we divided each pentagon into five triangles, but these were isoceles not right triangles. To use trigonometry we need right triangles, so we further subdivide each of these in two, giving ten right triangles.
PICTURE
If the side length of the pentagon is \(L\), then the outer side of each of these triangles has length \(L/2\). The two important angles here are \(\pi/5\) (the inside angle) and \(\pi/3\) (the outer angle that is not right). Thus, we see
\[\cos\frac{L}{2}=\frac{\cos\frac{\pi}{5}}{\sin\frac{\pi}{3}}\]
At this point it is totally reasonable to just calculate numerically via a calculator, and find that \(L/2 = 0.364\) so \(L=0.729\ldots\). However: we simplify further if we like, to see a relationship of the dodecahedron to the golden ratio! We know that \(\sin\pi/3=\sqrt{3}/2\) from the unit circle, but we also have
\[\cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}=\frac{\phi}{2}\]
Where \(\phi\) is the golden ratio. Thus, putting it all together
\[\cos\frac{L}{2}=\frac{\phi/2}{\sqrt{3}/2}=\frac{\phi}{\sqrt{3}}\]
and
\[L=2\arccos\left(\frac{\phi}{\sqrt{3}}\right)\]