The "inscribed angle theorem" is a common 2-dimensional plane geometry fact. It states that for a circle the angle formed between any two points on the circumference with the center is twice the angle formed by those two points with any other point on the circumference. I will not elaborate on a proof or further details here, but instead provide a link and image from the wikipedia page on this topic, where the basic theorem is proved.

IMAGE: The inscribed angle θ is half of the central angle 2θ that subtends the same arc on the circle (magenta).

Wikipedia Article on inscribed angle theorem

My question is whether this simple 2D geometric concept can be adapted to solid angles in 3D? And perhaps beyond to n dimensions?

The 2D case dealt with 3 points on the circumference of a circle (2 that defined the arc, and the 3rd point that formed the angle that was half the angle at the center). In 3D, imagine a sphere rather than a circle, and consider 4 points instead of 3. Let 3 of the 4 points form the base of a tetrahedron, then consider two distinct cases.

In the first case the 4th point forms the tip of the tetrahedron. In the other case the center of the sphere defined by the 4 points forms the tip of the tetrahedron. In either case the base of the tetrahedron, and the associated spherical triangle on the surface of the sphere, is the same. However there are two different solid angles at the tip of the tetrahedron in each case -- one for when the tip is at the center of the sphere and the other when the tip is at the 4th point defining the sphere.

Are these solid angles related (one being half of the other, or some other similar relation) as in the inscribed angle theorem in 2D? If they are related in some way, is this a common fact/theorem in solid geometry? Does it have a name like the "inscribed angle theorem" in 2D? What is the relation between these two solid angles?

Is there a similar concept in 4D, or n-dimensional, space? (I am not even sure if there is a solid angle concept in arbitrary n-dimensional space.) If there is a concept of solid angles in higher dimensions is there a predictable relation between these two angles a set n-dimensional space?For example, a particular dimension like 10-dimensional space, then 11 points in that 10D space, is there a way to easily find the "solid angle" at the 11th point, and then use a fixed relation to know the "solid angle" at the center of the 10D hypersphere going out to the other 10 points on the surface of the hypersphere?

  • Brown $ \theta$ is correct? – Narasimham Mar 24 '16 at 3:11
  • Narasimham, yes the brown/gold theta in the linked image is correct. – user669487 Mar 24 '16 at 5:07
  • Phew! for cycl. qudrilateral the opp angles sum to 180 deg – Narasimham Mar 24 '16 at 9:49

The answer to your question is no. The inscribed angle theorem does not work for inscribed solid angles in a sphere.

To see why, I will start by stating an equivalent version of the inscribed angle theorem. Consider the following figure.

enter image description here

This figure shows the point projection of a circle $C_1$ onto a circle $C_2$ of twice the radius. The projection point $P$ lies on the circumference of $C_1$ and is the center of $C_2$. In this situation, the inscribed angle theorem can be stated as follows:

The projection from $P$ of $C_1$ onto $C_2$ is length-preserving.

Note that this projection maps $C_1$ onto the lower half of $C_2$. The figure above shows two red arcs that have the same length under this projection.

So the proper question here is whether an analogous statement is true for spheres. Imagine two spheres $S_1$ and $S_2$, where $S_2$ has twice the radius of $S_1$, and $S_1$ is tangent to $S_2$ on the inside. If $P$ is the center point of $S_2$, then the right question to ask is the following:

Is the projection from $P$ of $S_1$ onto $S_2$ area-preserving?

Since doubling the radius of a sphere quadruples its area, this is the same as asking whether a solid inscribed angle is equal to one quarter of the corresponding central angle.

Note that this is certainly true infinitesimally near the point of tangency. That is, a solid inscribed angle that lies within an $\epsilon$-neighborhood of a diameter of a sphere is approximately 1/4 of the corresponding solid central angle.

Unfortunately, the answer to this question is no. This involves a simple calculation. Let the two spheres be $$ x^2+y^2+z^2=1,\qquad x^2+y^2+(z-1)^2 = 4. $$ So $P = (0,0,1)$ and the point of tangency is $(0,0,-1)$. We can parametrize the first sphere $S_1$ using spherical coordinates: $$ \Phi(\theta,\phi) \,=\, (\cos\theta\sin\phi,\,\sin\theta\sin\phi,\,\cos\phi). $$ It is easy to check that the projection from $P$ onto $S_2$ maps the point $\Phi(\theta,\phi)$ to the point $$ \Psi(\theta,\phi) \,=\, \bigl(2 \cos \theta \cos(\phi/2), 2\sin \theta \cos(\phi/2), 1-2\sin(\phi/2)\bigr) $$ on $S_2$. But $$ \left\|\frac{\partial \Phi}{\partial \phi} \times \frac{\partial \Phi}{\partial \psi}\right\| \,=\, \sin \phi $$ and $$ \left\|\frac{\partial \Psi}{\partial \phi} \times \frac{\partial \Psi}{\partial \psi}\right\| \,=\, 2\cos(\phi/2). $$ These are not the same, so the given map is not area-preserving. Indeed, since $$ \frac{2 \cos(\phi/2)}{{\sin\phi}} \,=\, \csc(\phi/2) $$ area is locally being multiplied by $\csc(\phi/2)$. As expected, this is equal to $1$ at the point of tangency (where $\phi=\pi$), but approaches infinity as we move toward $P$ (with $\phi\to 0$).

Practically speaking, what this means is that, given a region $R$ on a sphere and a point $P$ on the sphere that does not lie in $R$, the ratio $$ \frac{\text{central solid angle for }R}{\text{inscribed solid angle from }P\text{ to }R} $$ is close to $4$ when $R$ is small and diametrically opposite the point $P$, but approaches infinity when $R$ is small and very close to $P$. More specifically, the above considerations show that this ratio is the average value over $R$ of $4\csc(\phi/2)$, where $\phi$ is the central angle from $P$ to a point in $R$.

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