# Tag Info

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Even more is true for this theorem. Check out this drawing from Arseniy Akopyan wonderful book of Geometry in Figures (Second, extended edition, 2017). On page 65 we find Figure 4.7.29) In the foreword, Arseniy Akopyan writes "It is commonly very hard to determine who the author of a certain result is." He nevertheless provides source for many of the ...

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"$65537$-gon" is the name. Likewise "$257$-gon": writing (let alone saying) something like "diacosipentacontaheptagon" serves less to communicate $-$ if indeed it succeeds in communicating at all $-$ than to show off.

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Q1: Yes. Any acute non-isosceles triangle can be tiled by three pairwise incongruent isosceles triangles, by connecting each vertex to the circumcenter. Start from some isosceles $T_0$ with repeated side $s$; inscribe $T_0$ into a larger triangle $T_1$ such that $T_1 - T_0$ is the union of three acute, non-isosceles triangles with circumradii distinct from ...

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Here is a partial answer: in $\mathbb R^3$ there are cycles of length $2$. To show this, it is enough to find non-trivial systems of unit vectors $x_i, y_j$ ($i,j=1,\dots,d+1$) such that $(x_i,y_j)=t$ for some $t>0$ and all $i\ne j$. If $d=3$, we can take the normalized copies of $(1,0,4),(1,0,-4),(-1,4,0),(-1,-4,0)$ and $(-2,0,-1),(-2,0,1),(2,-1,0),(2,1,... 23 Let me elaborate on Sam Hopkins' comment. The main reason that makes this and other problems on continuous curves so hard is that a "simple closed curve" or "Jordan curve", i.e. a non-self-intersecting continuous loop in the plane, can be a horrible object, for instance a nowhere differentiable curve such as the Koch snowflake and other fractal curves. ... 23 (Initial post November 24, 2016, edited November 27, 2016) This does not exist. The proof that$X$doesn't exist is a bit elaborate and makes use of ends of coset spaces. I will prove: (a) Let$\Gamma$be a finitely generated subgroup of the group of isometries of the plane. If$\Gamma$is not virtually infinite cyclic, then every$\Gamma$-commensurated ... 22 There are several well-known and hard problems in discrete geometry that concern the dimension of a graph. For example, the unit distance problem asks for the maximum number of edges of a graph on$n$vertices of dimension 2. Erdős conjectured in the 1940s that the answer is$n^{1+o(1)}$, but the best known bound is only$O(n^{4/3})$. The chromatic number ... 21 Isn't this easy by induction? Delete one of the largest numbers, say$m$, from your sequence, realize the remaining numbers, then in the realization pick any region with winding number$m-1$, and make an extra loop somewhere close to its boundary. 19 Your question amounts to treating construction problems in geometry as decision problems, and so it makes sense to me to adopt the terminology of computability theory. This same kind of distinction arises in computability theory, where we have the following terminology: A set$A$is decidable if we can computably verify yes-or-no whether a given input$a$... 18 This is really more of a response to the OP's request for an 'insightful way to view the computation' than it is an answer to the specific problem; so keep this in mind. On the other hand, as I'll point out later, the features of this answer do generalize to handle other pairs of convex bodies, even if they can't be made as explicit, so I think it does shed ... 17 Google soon finds that Q2 is problem C11 in Unsolved Problems in Geometry by Croft, Falconer, and Guy. But perhaps it's been solved during the intervening decades. URL 17 Here's the abstract of K.A. Post, "Triangle in a triangle: On a problem of Steinhaus", Geom Dedicata (1993) 45: 115; this paper was cited in the one given in the comment by Nemo. A necessary and sufficient condition on the sides$p, q, r$of a triangle$PQR$and the sides$a, b, c$of a triangle$ABC$in order that$ABC$contains a congruent copy of$PQR$... 16 There is a set$P$. For the construction of this set first take the squares of area$1$whose edges are integers and numerate them. For each square, say$S_n$, you can take a square lattice in it such that any convex inside the square that do not intersect the lattice has area less than$a_n$for any$a_n>0$, just take a lattice with a very small distance.... 16 Let$E$be a set of the claimed form. Call a direction$\omega \in S^1$a limit direction of$E$if there exists a sequence$p_n$of points in$E$going to infinity whose argument goes to$\omega$, or equivalently if$E$does not avoid an infinite open sector containing the direction$\omega$in the limit. I can show the following: Proposition. The set ... 16 Simpler construction: a non-isosceles right triangle$T_0$can be divided into two isosceles triangles not congruent to each other, or into two right triangles similar to$T_0$. Reversing the latter construction, let$T'_n$,$T_n$($n = 1, 2, 3, \ldots$) be right triangles similar to$T_0$such that$T_n = T_{n-1} + T'_n$, and divide$T_0$and each$T'_n$... 16 @YosemiteStan's example. Detail: Tilt angle$=\sin ^{-1}\left(5 \sqrt{\frac{2}{61}}\right) \approx 65^\circ$. 15 Disproved by Viktor Kiss and Zoltán Vidnyánszky, see http://arxiv.org/abs/1402.5452 15 Not an answer, just a long comment. This comment slowly evolved into a partial numerical solution: the regular pentagon is the local maximum. I've used Maple to produce the formula for$\frac{S'}{S}$and the result is a rational function on$\mathbb{R}^{10}$with both numerator and denominator of degree 12. The numerator has 11495 monomial terms while the ... 15 A two-point set cannot be$F_\sigma$, as Mohammad mentions in his question. Also, A two-point set cannot contain a dense$G_\delta$subset of an arc. This was proved by Gareth Davies in his thesis (Oxford, 2011), but I do not think he ever published this result. To my knowledge, no better results are known in the Borel-sets-shouldn't-work direction. In ... 15 Here is recent paper coauthored by Cox (whom Yoav Kallus cited), with a different focus: Headley, Francis, and Simon Cox. "Least-perimeter partition of the disc into$N$regions of two different areas." arXiv:1901.00319 (2019). arXiv abstract. Their first figure illustrates Yoav's points that the arcs "meet in threes at$120^\circ$, but not ... 14 OK - following the suggestion of Mariano Suárez-Alvarez, a moderator - I post my comment as an answer: If k points are mutually invisible in an n-sided poygon, then no two of them lie in the same triangle of a triangulation. Therefore$h(n)≤n−2$, and there is a simple example, generalizing your construction for$n=4$and$5$, of an$(n+2)$-sided polygon ... 14 All of these Machin-like formulas have proofs-without-words. First, notice that there is a one line proof using the Gaussian integers: $$(3+i)^2 (7+i) = 50 + 50 i.$$ Taking arguments of both sides proves the result (modulo$2 \pi$). Now, plot the products $$3 \times 3 \times 7=63,\qquad (3+i) \times 3 \times 7=63+21 i,$$ $$(3+i)^2 \times 7 = 56+42i,\qquad (... 14 Illustration of Noam's construction: 14 I don't think there is such a set for the plane, but I'll point out that there is one for the sphere \mathbb S^2. Namely, let S and T be two members of SO_3 such that the group G they generate is free. Take any x, y \in \mathbb S^2 such that y \notin G(x). Each member of G can be uniquely written as a "reduced word" U_1 \ldots U_n where ... 14 Here are a few trivial lemmas. I won't use anything about the rolling motion, just that the distance is defined by gluing pentagons edge-to-edge: The dd-circle of radius k, which I'll call C_k, is a closed polygonal curve. Let D_k be the closed dd-disk of radius k; note that D_k may contain holes and C_k is not in general simple! (This ... 14 The number is tabulated at OEIS. It seems that it's only known up to n=14 (and some scattered larger values). Links are given there to some papers on the topic. Evidently, no one knows how to do it for general n. Also discussed on math.stackexchange. 14 Yes, \arcsin(\frac14)/\pi is irrational. Suppose \arcsin(\frac14)/\pi = m/n, where m and n are integers. Then \sin(n \arcsin(\frac14))=\sin(m \pi)=0. We analyze this usng the formulas from Browmich as cited on Mathworld:$$\frac{\sin(n\arcsin(x))}{n}=x-\frac{(n^2-1^2)x^3}{3!} + \frac{(n^2-1^2)(n^2-3^2)x^5}{5!} + \cdots\frac{\sin(n\arcsin(x))}{... 13 [Corrected (two typos noted by V.T.) and expanded (alternative coordinates)] Also not an answer, but this time a simpler algebraic formulation. All indices are cyclic mod$5$. Let the vertices be$P_i$($i \bmod 5$), and let$A_i$be the area of triangle$P_{i-1} P_i P_{i+1}$. Then$S$is the larger root of the quadratic equation$$S^2 - S \sum_{i\,\bmod\... 13 The answer to the question is no. Rather than typing a new answer, I've just edited my old one (which contained some partial progress). I think this is OK since the main idea of the old answer (or at least its main "trick" -- fixing a special line, and then rotating it slightly about a point not in$E$) is still present in this one. The proof I'm about to ... 13 On the second question: There is the inequality$12\sqrt{3}A\leq P^2$that needs to be satisfied first of all to get a triangle. Using Heron's formula for a triangle with sides$x$and$y$, you are asking for the rational solutions to$(x+y-P/2)(P/2-x)(P/2-y)=2A^2/P$. The projective closure of this curve is an elliptic curve with$O=(1:-1:0)\$ and two other ...

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