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Let me give a proof of rozu's guess that $\phi_3(D)=1/2$ and $\phi_3(S)=\sqrt{2}-1$ if $D$, $S$ are a disk and a square. This disproves your conjecture. First of all, for every $K$ and every $N$ the c …

Your argument already gives $$2\cdot\frac{(n-6c)^3}{216}+3\cdot \frac{(n-6c)^3}{108}=\frac{(n-6c)^3}{27}$$ triangles.

"Only if" part (for a symmetric region, bisectors are concurrent) is rather clear. Now "if" part. Assume that the common point of all bisectors exists, denote it by $O$. Clearly every chord $AOB$ thro …

Is not it obvious (unlike Prékopa-Leindler)? We are given that for all $x_1,x_2$ we have $(f_1/g_1)^2 (x_1)\leqslant (g_2/f_2)^3(x_2)$, thus there exists $c>0$ such that $(f_1/g_1)^2 (x_1)\leqslant c^ …

Does not an elementary Euclidean proof work verbatim? If both sets are polygons, the interior one is obtained from the exterior one by cuts (i.e. transformations $P\to P\cap H$, where $P$ is a polygon …

Let $A_0\subset A$ be the set where $\ell$ attains a minimum on $A$. It is a face of some dimension $k<d$. If $k=d-1$, we are done. Assume that $k<d-1$. Without loss of generality, $0\in A_0$ and more …

I do not think so. Consider a uniform matroid on 3 elements $a, b, c$ of rank 2, and take 100 copies of $a$ (so, totally we have 102 elements). Then the matroid base polytope has full dimension 101, t …

A planar convex region is centrally symmetric if and only if its area bisectors are all concurrent. Yes. Let all area bisectors of our region $K$ pass through a point $O$. Note that any line $\ell$ …

Looks true. A necessary and sufficient condition for these points (let $E$ be a standard ball) is that the identity operator $I$ is a non-negative linear combination of projectors $P_i$ on lines throu …

Yes, this is true in general case. Let $P={\rm conv} M$ for a finite set $M\subset \mathbb{Q}^n$, and let $x_0$ be a minimizer of $\|x\|^2$ over $x\in P$. I am going to prove that $x_0\in \mathbb{Q}^n …

Assume that $q\in C_2\setminus C_1$. Let $p$ be an interior point of $C_1$. Then the interval $(p,q)$ contains a boundary point of $C_1$ but only interior points of $C_2$. A contradiction.

If $f(x)=\min\{s>0:x/s\in B\}$ is Minkowski functional of $B$, then $f$ is a convex function on the plane and ${\bf p}(t)=\frac{e^{it}}{f(e^{it})}$. I think your claims now follow from the properties …

If $p_1,\dots,p_{n+2}$ are the vertices and $\sum \lambda_i p_i=0$, $\sum \lambda_i=0$, is their (let it be unique for a moment) affine dependence, the point $q=\alpha p_i+(1-\alpha) p_j$ on the diago …

I do not know the reference, but it looks that even more may be achieved: $t_1,\dots,t_{n-1}$ may be chosen close to 0 and $t_n$ close to 1. Indeed, if the span of $a\cup A$ is spanned by linearly ind …

This follows from the Lemma. If $F$ is a face of $conv(X)$, $x\in F$ and a finite subset $A\subset X$ is inclusion-minimal subset for which $x\in conv(A)$, then $A\subset F$. Proof. Induction in $|A …