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I took a pane of clear glass and touched two balls at once I put my light, perhaps, by chance, above the pane. Alas, the shining pile on the same side in its arrangement lay, and no matter what …

That isn't a conjecture but a routine exercise assigned after the students learn about Bang's solution of the Tarski plank problem. The proof goes in 2 steps: 1) Consider all sums $\sum_j \varepsilon …

Zachary Chase said that he is still interested, so I'll make a post. It is a bit on the long side and I'll need to draw a few pictures to make it comprehensible, so today I'll just post a counterexamp …

Anyway, If you know a 5 line proof for the first inequality please share it with us OK, here goes. Assume that $\inf_P f=-1$ and that it is (nearly) attained at the origin. Let $K=\{x\in P: f(x)\le …

I would try to approach your original problem a bit differently. Note that $|x-a|\le \frac 12(r|x-a|^2+r^{-1})$ and the equality is attained for $r=|x-a|^{-1}$. Thus, $$ \min_x[|x-a|+\langle b,x\rang …

Here is a cheap probabilistic example of a symmetric convex body with the minimal projection greater than the maximal section. Take the unit ball (say, in $\mathbb R^3$) and remove $m$ pairs of symmet …

Consider $f(x,y)=x^3+ay^3$ ($a>0$) on $K=\{x,y\ge 0,x+y=1\}$ (or in a thin convex set around this interval). The minimum of $f$ is at the point where $x^2=ay^2$ or $x=a^{1/2}y$. The square of the grad …

Let $\sigma$ be represented by a PD matrix $A$ and $\rho$ by $A+B$. Note that $|B|\ge B$ (in the sense of PSD matrices) and has the same $1$-norm. Also $\Pi |B|\Pi\ge \Pi B \Pi$, so to dominate $\Pi B …

Yes, because a connected graph ($i\sim j$ iff $\angle x_ix_0x_j$ is more than $\frac\pi 2$) with $n+1$ vertices has at least $n$ edges. The graph is connected because (assuming WLOG that $x_0=0$ is st …

In other words, you want the set of points $d$ such that if you rotate the plane $bcd$ so that it makes one plane with $abd$, then $ab$ and $bc$ make one line. Since you know that line, the question b …

To those who are still interested: we've finally made it but it's so ugly that a nice alternative approach will be always welcome :) We are still stuck with Bonnesen's question about the possibility t …

Yes, of course. Claim 1: the VC dimension of translations of a fixed triangle on the plane is at most $3$. Proof: Take any $4$ points consider the $3$ points lying farthest in the directions of the …

A full answer was given in the comments. It is (almost) exactly the same argument as in the linked thread, so the only purpose of typing this "answer" is to remove the question from the "Unanswered" l …

OK, here is why $Lip1$ is impossible. Suppose we have such choice. Then consider all $K$ whose minimal containing box is a unit square. Let $[x(K),x(K)+1]\times[y(K),y(K)+1]$ be the minimal containing …