# Tag Info

0

Bringing together complex numbers and laziness. Following the answer by @FedorPetrov, we see that $N=(A+B+C)/2$ and so $A'=B+C$. Take inversion with respect to the unit circle (which is just a map $z\mapsto 1/\bar{z}$), then the circles are mapped to the lines connecting $A$ and $BC/(B+C)$ (and two analogous) and we want to prove they intersect at one point. ...

3

Tait–Kneser theorem says that generically curves crosses its osculating circle. If not (that is, if the curve is locally supported by its osculating circle), then the point is a vertex of the curve, but the supporting condition is stronger a bit. If it is a point of local minimim/maximum of the curvature, then the curve lies locally outside/inside of its ...

1

Assuming the curve $\gamma(s)$ is parametrised in arc length, if for a disk $B(p,r)\subset\mathbb R^2$ one has $\gamma(s_0)\in\partial B(p,r)\subset\mathbb R^2$, then $\gamma(s)$ is locally outside the disk $B(p,r)$ at $s_0$ iff the distance $\|\gamma(s)-p\|^2$ has a local minimum at $s=s_0$. For the osculating circle at $\gamma(s_0)$, that is $p=\gamma(s_0)+... 1 This is a short answer to Question 3 above: Theorem. Every continuous symmetric distance function on the$n$-sphere for which great circles are geodesic can be obtained as a limit of linear combinations of distance functions of the form $$d_A(x,y) = \arccos\left(\frac{Ax \cdot Ay}{||Ax|| ||Ay||} \right) \ ,$$ where$A$is an invertible linear map from$\...

5

It is $\sqrt{2}$. For 4 vertices of a square, you get this value. For proving that it is always not less than $\sqrt{2}$, note that one of angles $\angle A_iA_jA_k$ is not less than $\pi/2$ (if $A_1A_2A_3A_4$ is a convex quadrilateral, the sum of angles equals $2\pi$, thus one of them is at least $\pi/2$; if $A_4$ lies in a a triangle $A_1A_2A_3$, the sum of ...

9

A curvature bound for a non-smooth metric space is often known as a "synthetic curvature bound" or a "coarse curvature." Here's one possible definition for the concept. Definition: A condition $Q_{\kappa}$ is a synthetic lower bound for a curvature tensor $R$ if the following two conditions hold: (1) On a smooth manifold $M$ where $R$ is ...

3

Not an answer. I'm just expanding a comment about @PeterTaylor's observation that the known pseudovertices $X(4)$, $X(74)$, $X(1138)$ lie on the Neuberg cubic ... Bernard Gibert's "Pairs and Triads of points on the Neuberg Cubic connected with Euler Lines and Brocard Axes Isometric Parallel Chords" Proposition 1 characterizes the Neuberg cubic of $\... 4 This is a report on an unsuccessful computational approach which is rather too long for a comment. I work with complex numbers to represent the points in the obvious way. It suffices to consider$\mu(z) = t(z,0,1)$because this can be extended under the invariants to the full$t(z,z',z'')$. Since multiplication by a complex number is just rotation and ... 11 These are called halving lines, and we don't know the exact order of their magnitude, just that it is between$\Omega(n2^{\sqrt\log n})$and$O(n^{4/3})$. For more information, see https://jeffe.cs.illinois.edu/open/ksets.html. 2 By the Wallace-Bolyai-Gerwien theorem it suffices to cut the polygon into$n$sets of equal area, which can certainly be done by continuity of the area on one side of a line as you move the line across the shape. If any of the sets is discontiguous you can rearrange pieces to make it contiguous. Then apply WBG to rearrange each set into the shape of an ... 2 Every$n$is possible. As you point out, it suffices to answer the question for triangles. You can divide a triangle$T$into$n^2$congruent triangles similar to$T.$Then these can be partitioned into$n$sets of$n$which are congruent as sets and, indeed, have all members congruent. It is interesting to note that, If you triangulate an$m$-gon$P$into$...

6

The OP doesn't say what is meant by a 'geometric object', so it's hard to give a definitive answer. However, if one assumes that the geometric object is a smooth manifold $M^7$ and that the action is smooth, then there are a few things one can say: First, the $\mathrm{G}_2$-stabilizer of any point $p\in M^7$ has to be a closed subgroup $H_p\subset \mathrm{G}... 2 Your statement is a pointwise statement. It essentially follows directly from the definition of a "Clifford multiplication". To understand it, you first have to be aware of the fact that there are two conventions for the Clifford relations in the literature: 1)$XY+YX+2<X,Y>=0$and 2)$XY+YX-2<X,Y>=0$. It seems that you use convention 2)... 1 Let$R\subseteq P$be the region$\{(x,y)\in P:x>0,y>0, x+y>1\}$and let$g:P\to R$be continuous and bijective. Let$h(p_2, p_3, p_4; (x, y)) = p_3+(p_2-p_3)y + (p_4-p_3)x$. Note that$h: P^4 \to P$is continuous, and that$h(p_2, p_3, p_4; q_i)=p_i$, where$q_2=(0,1)$,$q_3=(0,0)$, and$q_4=(1,0)$. Let$f_1(p_1,p_2,p_3,p_4) = h(p_2, p_3, p_4; g(...

0

If the object does not have an isotropic moment of inertia, then its symmetry group must be reducible in the language of group representations, i.e., a direct sum of lower-dimensional representations. That is, it must be a subgroup of O(1)×O(3) or O(2)×O(2), or even of O(1)×O(1)×O(2) or O(1)×O(1)×O(1)×O(1). This is easy to see by looking at the ellipsoid E ...

0

An asymptotic reformulation of this conjecture has now been solved by Diamond and Yehudayoff in the paper Explicit Exponential Lower Bounds for Exact Hyperplane Covers. Preprint is available here. The sharp form of the conjecture is still open.

19

Using the tiles suggested in Matt F.'s answer, there is a solution for the regular $6$-gon:

3

The quadrilateral $((0,10),(0,0),(6,0),(3,6))$ is an example where the minimum-area triangle (in blue) is distinct from the minimum-perimeter triangle (in red). The analysis for it shows how to answer the full title question numerically. We use the following results. Lemma for Area: If triangle $ABC$ has minimal area among triangles enclosing a convex ...

3

My personal favorite proof is described well in this blog post (which attributes it to Hermann Karcher, though I first heard a version of it back in graduate school, so it should probably just be called folklore). It is entirely synthetic and calculation-free.

1

Section 7.19 (Hexagons) of Beardon's book The geometry of discrete groups gives a proof.

Top 50 recent answers are included