Let $H,H'\subset\mathbb{R}^m$ be two hyperplanes with unit normal-vector, and let $P\subset\mathbb{R^m}$ be a convex polytope (defined via its corners $v_0, ... , v_n$, where $n\ge m$).

Let's further assume that

  • $dist(v_i,H) = 0,\ i\in [0,k]\ \wedge\ dist(v_i,H) \gt 0,\ i\in[k+1,n]$

  • $dist(v_j,H') = 0,\ j\in [0,k+d]\ \wedge\ dist(v_j,H') \gt 0,\ j\in[k+d+1,n]\ \wedge\ d\ge 1$

Here $dist(\ ,\ )$ denotes the signed distance.


Is it true that under the conditions stated above, we have $$ \sum_{i=0}^{n}dist(v_i,H)\ \gt\ \sum_{j=0}^{n}dist(v_j,H'), $$
respectively, are there counterexamples known?

As Fedor Petrov's shows, that is not generally true, but (as a follow up question) what about $$ \sup_H\sum_{i=0}^{n}dist(v_i,H)\ \gt\ \sum_{j=0}^{n}dist(v_j,H') ? $$

  • $\begingroup$ You intend $k$ to be fixed? In what range? $0 \le k < n-d$? $\endgroup$ Commented Oct 8, 2015 at 21:15
  • $\begingroup$ $k$ denotes the number of vertices defining a hyper facet of the convex polytope or the polytope itself. The question amounts to whether the above defined distance sum decreases if additional corners are brought into contact with the hyperplane, i.e. if the contact Dimension is increased. $\endgroup$ Commented Oct 10, 2015 at 3:08

2 Answers 2


No. Take a triangle $v_0v_1v_2$ on the plane, $H'$ is its side $v_0v_1$, $H$ is almost another side $v_0v_2$. Then $\sum_{i=0}^2 {\rm dist}\, (v_i,H')$ is just a length of altitude from $v_2$, $\sum_{i=0}^2 {\rm dist}\, (v_i,H)$ is almost the length of altitude from $v_1$, which may be less than that from $v_2$.

As for your new question, the answer is still negative for triangles. Assume that a median $v_0 p$ is perpendicular to $H'=v_0v_1$. For any line $H$ passing through $v_0$ and not cutting the triangle we have $\sum_{i=0}^2 {\rm dist}\,(v_i,H)=2\,{\rm dist}\, (p,H)\leqslant 2|v_0p|$ and maximum is achieved for $H=H'$.

  • $\begingroup$ did you mean "both sums" instead of "both sides"? $\endgroup$ Commented Jan 16, 2016 at 12:43
  • $\begingroup$ Both sums (or both sides of inequality). $\endgroup$ Commented Jan 16, 2016 at 13:01

It just occured to me that the question can be answered in the affirmative way for the following reasons:

  • the order relation doesn't change if the distance sum is divided by some positive constant value $c$.

  • if we divide the distance sums by $n+1$, then that value equals the distance of the vertices' center of mass $R$ to the same plane.

  • all local maxima of $R$'s distances to points on the convex hull of the vertices are attained in a vertex which also is a corner of the convex hull and consequently also the global maximum of the distances is attained in such a vertex.

  • $R$'s distance to a plane $E$ containing a vertex $v$ that also is a corner of the convex hull attains a local maximum if $R-v$ is orthogonal to $E$.

  • A vertex $v^*$ with maximal distance to $R$ lies on the boundary of the smallest hypersphere centered at $R$, wich contains all vertices and, the plane $E^*$ that contains $v^*$ and is orthogonal to $R-v^*$ consequently also is a tangent plane of that hypersphere and thus the intersection of $E^*$ with the vertices' convex hull is $0$-dimensional and identical to $v^*$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.