# A trigonometric problem when calculating distance to the boundary of a convex hull

Suppose we have a sphere and a point outside of the sphere.

We denote the point outside as $v$ and the origin of the sphere as $x$. The convex hull of the sphere and $v$ should be like an ice cream cone.

So for a point inside of the convex hull, say $u$, I want to prove that the distance to the boundary of the convex hull can be solved by a reduced problem where we consider only the part of the plane intersecting the convex hull containing points $x, u, v$. The reduced problem is illustrated in the figure.

Basically, the question is: Does this restriction to the plane give the shortest distance to the boundary of the convex hull? How can we prove it?

This seems like a trivial fact but it does not seem like that easy to prove...

From any other ray propagated from $v$ on the convex hull, we draw a segment from u perpendicular to it. Let the distance between $v$ and $u$ be $L$, and the angle between $\overline{vu}$ and the two rays be $\alpha$ and $\beta$, respectively. Since $\alpha$ and $\beta$ must be in $(0,\frac{\pi}{2})$, and length of the two distances we need to compare is $L\sin{\alpha}$ and $L\sin{\beta}$. It would suffice just to compare $\alpha$ and $\beta$. But I don't know how to proceed from here.

Any suggestions would be appreciated.

• The convex hull is only a cone if you are thinking of an ice cream cone. Jul 20, 2016 at 4:23
• Yes....That is what I meant.....
– jack
Jul 20, 2016 at 4:33
• What does transverse mean?
– zhw.
Jul 22, 2016 at 15:42
• It is the plane that intersects the convex hull and passes through $x,u,v$....
– jack
Jul 22, 2016 at 15:44
• I do believe the simplest argument would start with the observation that both $x$ and $v$ are on the axis of symmetry (by definition -- the former is the center of the sphere, and also on the axis of the right circular cone, with $v$ being the apex of said cone). Jul 23, 2016 at 13:35

The image on the left is a side view and the image on the right is an end view along the $x$-$v$ axis.

The circle in the right image is the intersection of the dotted plane in the left image with the convex hull, which is a surface of revolution around the $x$-$v$ axis, which means that each cross section perpendicular to the $x$-$v$ axis is circular and centered on the $x$-$v$ axis.

The distance from $u$ to the points on the convex hull, is $\sqrt{h^2+r^2}$. As can be seen in the right image, $r_{\text{min}}$ is in the plane containing $u$, $v$, and $x$.

Therefore, the point with the minimum distance is in the plane containing $u$, $v$, and $x$.

That is, the restriction to this plane does give the shortest distance to the boundary of the convex hull.

• Perhaps it would make it clearer if you added, after mentioning the convex hull being a surface of revolution after the $x-v$ axis, that it also means the convex hull is radially symmetric with respect to the axis through $x$ and $v$? Jul 23, 2016 at 13:37
– robjohn
Jul 23, 2016 at 13:51
• Yes, that works quite well now; +1. The key to understanding the situation is the axial symmetry -- your illustration with the cross section makes that clear now, I think. I am terrible (non-operational, really) at proofs, so that aspect I have no opinion on; but certainly now the geometric basis for the proof should be obvious to non-maths people like me, too. Jul 23, 2016 at 15:50
• Are those points co-planar if we replace the sphere with 4-sphere or any other general n-sphere?
– jack
Sep 3, 2016 at 23:28
• @jack: $v$, $x$, and $u$ are three points. Three points determine a $2$-dimensional plane in any dimension.
– robjohn
Sep 4, 2016 at 8:07

I'm not entirely sure what you mean by the transverse of the convex hull, but assuming this is equivalent to the plane defined by $u$, $v$, and $x$, I believe the following approach leads to a relatively simple proof:

We can find the shortest distance by expanding a sphere outwards from $u$ until it is tangent to some point $t$ (if there are multiple such points, choose one) on the boundary of the cone. That this point of contact is in fact a point of tangency follows from the convexity of the region, I believe.

Then we know that $\overline{ut}$ will be orthogonal to the plane of tangency at $t$. But we also know that any such lines orthogonal to the region at some $t$ will intersect $\overline{vx}$, from the basic geometry of cones and spheres (by symmetry, all such lines pass through the axis of rotational symmetry of a cone and through the center $x$ for the sphere). From the fact that $\overline{ut}$ and $\overline{vx}$ intersect, we have that they are coplanar, which is what we set out to prove.

The problem is not connected to the presence or absence of the sphere, as we are not concerned with the circular part. The cone has the convexity.

After recognizing that we are restricted to the central plane due to cone symmetry we need to consider only the central section containing cone axis.

Take $u$ as origin. If $p$ is minimal normal distance to cone generator which has inclination $\pi/2 + \alpha$ to $v-x$ axis of symmetry, then cone generator as $x \cos \alpha + y \sin \alpha = p$ has minimal normal distance.

Putting it another way..Let vector perpendicular to the generator and line through $u$ be the axis $X$ considered out of the plane. An intersection of a plane through $X$ and $u$ to cone is an ellipse. The minimum geodesic curvature of ellipse intersection occurs at its major axis by Meusnier's theorem:

$$R_{slant} = R_{normal} \cos \varphi$$

where $R_n$ is the largest radius of curvature among all inclinations of a sphere of diameter $R_n$ at $\varphi =0.$ It is easily seen that only the normal section has maximum radius of curvature. All rotations of a circular arc around shortest normal as radius / polar axis do not cut but are tangential to cone.