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Hilbert's axioms predate the development (or at least the wide adoption) of fully symbolic logic, so they are expressed in partially informal language -- though Hilbert strove to make them as precise as he could. They include the Axiom of Archimedes, formulated in language that presupposes that the natural numbers are already known. As such, if we want to ...

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You may assume the first point $A$ at $(1,0)$ and the second point $B=(\cos\phi,\sin\phi)$ being uniformly distributed on the circle. The probability measure is then given by ${1\over2\pi}{\rm d}\phi$. The distance $D:=|AB|$ computes to $2\left|\sin{\phi\over2}\right|$, and we obtain $${\mathbb E}(D)={1\over 2\pi}\int_{-\pi}^\pi ... 5 The reason I think is simplicity:$$A=\sqrt{s(s-a)(s-b)(s-c)}$$Is much more simple than:$$A=\frac{1}{4}\sqrt{(a+b+c)(b+c−a)(c+a−b)(a+b−c)}$$I see in the comments that you say we have to add \text{where } 2s=a+b+c, in the standard form, and so the length of the two forms become the same. However, this is a useless argument. Continuing with the logic ... 2 Question Every edge of a triangle contains all but one of the vertices. Every face of a tetrahedron contains all but one of the vertices. Every (n-1)-face of an n-simplex contains all but one of the vertices. Every edge of q square contains half of the vertices. Every face of a cube contains half of the vertices. Every (n-1)-face of an n-cube ... 2 Because Euclidean geometry is currently not fashionable, most people do not study topics in it or discuss problems in it, and so you simply hear of fewer problems, solved or unsolved. Any claims that all Euclidean geometry problems are decidable, as given in the comments to the question, will depend on some restricted definition regarding the form that a ... 1 Triangle inequality is sufficient enough. From \triangle ABD we get that$$BD > |AD - AB|.$$Both AD and AC equal to the radius of the circle so AD = AC, and inequality transforms into BD > |AC - AB|. I hope it is clear that |AC - AB| = BC. 1 It might be worth stating the theorem: for any finite set of points in the Euclidean plane either the points are collinear or else there is a line passing through exactly two of the points. This theorem was probably not proved with any particular applications in mind. Nevertheless, on my first page of Google results, I find this paper studying ... 1 Pez, if I understand correctly, you want to know how each polygon vertex v = (a_i,b_i) moves as you move each edge along the edge normal by \epsilon? First, compute the normals N_1, N_2 of the two edges adjacent to v. Then the inflated vertex position v' is given by$$v' = v + \frac{N_1+N_2}{\|N_1+N_2\|} \epsilon \sec \frac{\psi}{2},$$where ... 1 You need to see the proof of that formula. If the proof has not to do directly with the concept of semiperimeter, than it means that two equivalent ways writting that formula are of the same power (substantially speaking). One of the proves is in this link here: http://jwilson.coe.uga.edu/emt725/Heron/HeronProofAlg.html and it has to do nothing with the ... 1 It turns out to be false for general cyclic quadrilateral. Let the triangle have unit side length. Take point P on circumcircle so that PA=\frac{3}{7}; PB=\frac{5}{7} (it would lie on a circle, thats easy to prove from cosine rule). By Pompieu theorem, PC=\frac{8}{7}. Now take points Q,R,S with the same but permitted distances, and we have a ... 1 What you are talking about is the same as changing$$ e^{i\pi } + 1 = 0 $$to$$ e^{\frac{i\tau}{2}} + 1 = 0. $$The first one widely believed to be the most beautiful formula in mathematics. Although the may be the same, and although many schools use Tau in mathematics, you lose a sense of beauty and simplicity that you had in the first formula. In ... 1 Wikipedia explains the differences and similarities between the two. Finding Hermite normal form for an integer matrix, you're not allowed to divide, so you won't necessarily get a leading 1 in each nonzero row, for instance. 1 First find distance between all possible pairs of bikers and bikes. Now sort this distances. Now run a loop through all the distances, and create edges only between bikes and bikers whose cost is less than current distance you are considering. Now run Hopcroft-Karp algorithm on that sub graph and compare the max cardinality given by algorithm to K. If they ... 1 If a is even and coprime to b then it is represented by  2mn  for  gcd(n,m) = 1  and n , m not both odd in Euclids Formula. http://en.m.wikipedia.org/wiki/Pythagorean_triple  c + a = m^2 + 2mn + n^2 = (m + n)^2   c - a = m^2 - 2mn + n^2 = (m - n)^2  1 Hint: label the points on the polygon A,B,C,D,E,F such that \angle ABC = 36^{\circ} and \angle BCD = (360-84)^{\circ} = 276^{\circ}. For any polygon with n sides, the sum of the internal angles must be (180(n-2))^{\circ} = (36 + 276 + 4a)^{\circ}. Can you figure out why, and can you take it from here? 1 Let \{v_{i}\}_{i=1}^k be a set of k vectors in \mathbf{R}^n. By "sum of all pairs of inner-products", presumably you mean something like$$ \sum_{i<j} \langle v_{i}, v_{j}\rangle, $$and by "sum of Euclidean distances between all pairs" you mean$$ \sum_{i<j} \|v_{i} - v_{j}\|.  Consider what happens with two vectors. Since you're asking about ...

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In this context, "linear distance" appears to mean "arc length"[*], while "angular distance" is presumably $\Delta\theta$, the angular difference measured at the center of the circle. These quantities are indeed proportional, with the radius $r$ as constant of proportionality. [*] Note, for example, that arc length (rather than chord length) is the relevant ...

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