# Tag Info

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Why would you have to say “space exist” before you can say “points exist”? In order to have something that contains all those points? But then by the same reasoning, you can't say “space exist” without first postulating the existence of some entity that can contain spaces. Before you know it, you require an infinite regress of existence statments. In short, ...

6

Because Euclid axioms define a space.

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A more general approach is given as follows. Suppose we have the points $$\zeta_{n,0}^k = \left( \cos \frac{2\pi k}{n}, \sin \frac{2\pi k}{n} \right), \quad k = 0, 1, 2, \ldots, n-1$$ forming a regular $n$-gon with unit circumradius in the Cartesian coordinate plane. We seek a linear transformation $T$ that maps $\zeta_{n,0}^k$ to a point $\zeta_{n,1}^k$ ...

5

Let's say your original 4 points are, going counterclockwise from the origin, $$(0,0),(100,0),(100,100),(0,100)$$ You want to rotate clockwise, which means the point that was on the origin will go up to $(0,a)$ and the point that was at $(0,100)$ will go to $(a,100)$. And if you draw out this new situation with all 4 little wedges filling in the space ...

4

A space is a collection of objects that obey certain properties. That is, a space is a set with a structure. In this sense, Euclid's geometry doesn't need to assume the existence of a space, because he defines Euclidean Space. Consider also that Peano doesn't assume the existence of numbers, nor do Zermelo and Fraenkel assume the existence of ...

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Hint If $P$ is inside $\Delta ABC$, the angles must sum to $360^\circ$, if $P$ is outside, the angles can never be all equal. Can you now prove that $P$ must be unique? (It even has a special name)

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Note that $E$ and $I$ both lie on the bisector of $\angle C$, so $C$, $E$ and $I$ are collinear. Similarly $B$, $F$ and $I$ are collinear. Since $F$ lies on the angle bisector of angle $\angle BAI$, we have $\frac{|BF|}{|FI|} = \frac{|AB|}{|AI|}$ by the angle bisector theorem. In the same way, we have $\frac{|IE|}{|EC|} = \frac{|AI|}{|AC|}$. By the angle ...

3

In Euclidean geometry, you may refer to points, lines, and circles; implicitly and conceptually, these all reside within some "space", but Euclid's geometry doesn't include the vocabulary to refer to this whole space. Modern axiomatic mathematics certainly does have the capability to refer to the full entity of Euclidean space, naively represented as ...

3

Hint: Try to prove that $\Delta ABD \cong \Delta ACM$ using parts $(1),(2)$ of the problem.

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we will use that an arc makes same angle on the circumference. $\angle BMC = \angle BAC = 60^\circ$ angle made by arc $BC$ similarly $\angle BMA = \angle BCC = 60^\circ$ therefore $\angle BMC = \angle BMA$ and that shows $BM$ bisects $\angle AMC$ (2) look at the isoscles triangle $AMD$ with one base angle $AMD = 60^\circ$ that makes $AMD$ equilateral ...

3

$(3)$ is a consequence of Ptolemy's theorem, from which: $$MB\cdot AC = MA\cdot BC + MC\cdot AB.$$ Since $AC=BC=AB$, $$MB = MA+MC$$ follows.

2

Given the diagram as labeled ... ... we consider that construction occurred as follows: Starting with point $P$ (or point $Q$) on $\overrightarrow{OR}$, construct $\overleftrightarrow{AP}$ (or $\overleftrightarrow{BQ}$) and let $C$ be the point where this line meets the unit circle. Then $\overleftrightarrow{BC}$ (or $\overleftrightarrow{AC}$) ...

2

$AC \le AP + PC = 24 + 28 = 52$ and $BD \le BP + PD=32+45 = 77$ adding $AB + BC < AC \le 52, BC + CD < BD \le 77,CD + DA < CA \le 52$ and $DA + AB < DB \le 77$ we find that the perimeter is bounded above by $129.$ writing $AB = a, BC = b, Cd = c, DA = d,$ we have $$a + b + c + d \le 129, ac+bd \le 52*77=4004$$ i am stuck and not sure if ...

2

There is insufficient information to find the perimeter. Draw point P and make 4 line segments from it, of given lengths. There is no way of finding the angles between them, and there are many cases that give a convex quadrilateral.

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A straight parabola (i.e., one with its directrix parallel to the $\;y$- axis) has its maximal or minimal point at its vertex (depending on its leading coefficient's sign), and $$f(x)=y=(x-h)^2+k\implies f'(x)=2(x-h)=0\iff \color{red}{x=h}$$ and then $$f(h)=k$$ so the vertex is indeed at $\;(h,k)\;$ , and it is a minimum point iff ...

2

I think you gave a wrong hypothesis: the line passes by the point $(2,1,10)$ and parallel to the vector $(2,-3,8)^T$ and the plane passes by the point $(2,0,0)$ and parallel to the plane spanned by the vectors $(2,4,1)^T$ and $(-1,2,1)^T$. From this hypothesis we get the given system of equations.

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You need to use trigonometrical functions to normalize the angles which are discontinuous around the boundaries. It can be done using combination of sin and cos, arctan2 functions. I use following java code for filtering 9-axis IMU sensor values. I hope this helps. public class AngleSmoothingFilter { private FloatSmoothingFilter sinFilter = new ...

2

You have to have enough information to know what cycle you are in. The easiest way is to sample the data often enough that the change in angle is always well less than $180^\circ$ between samples. Then you can just add or subtract $360^\circ$ as required to make the minimum change and smooth after that.

2

Where is it located? Unfortunately I couldn't find a closed formula for the trilinears yet. Here is what I have so far. Let the outer triangle have edge lengths proportional to $a_1:b_1:c_1$. The trilinear coordinates of the recursive contact center with respect to that triangle would be called $x_1:y_1:z_1$. Now the edge lengths of the contact triangle ...

2

Here is an elaboration of the comment by user1551. I will prove, by induction on the dimension: If $E$ is a Euclidean vector space of dimension $n$, and $v_1,\ldots,v_k\in E\setminus\{0\}$ are such that $v_i\cdot v_j\leq0$ whenever $i\neq j$, then $k\leq 2n$. Taking the vectors of an orthogonal basis and their opposites, one sees that the bound can be ...

1

Let us just try to draw the figure. Since $AC$ bisects $\widehat{BAD}$, given that $B'$ is the symmetric of $B$ with respect to $AC$, we must have that $D$ lies on the $AB'$ line. Given that $l$ is the perpendicular bisector of $BC$ and $A'$ is the symmetric of $A$ with respect to $l$, in order to have $\widehat{ABC}=\widehat{BCD}$ we must have that $D$ lies ...

1

I believe you are going about this the wrong way. The axiom you are trying to say is that some space exists to contain some points. However, by the mere existence of the points, we already have a space. The space is the collection of points and what the axioms allow. Nothing more, nothing less. If you agree that there are a bunch of points, then simply call ...

1

Yes, you consider a transformation taking circles to circles ( or lines) and taking one of the points to infinity. The three circles become three lines through the image of the other point. Note that your transformation preserves the angles. Now it's easy. Exactly what you did.

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I found out that an answer to the first question (the total number of loops of a certain length) can be furnished by algebraic graph theory, and in particular the adjacency matrix. All one needs to do is to represent any given configuration of polysticks as a graph $G$ and then form the adjacency matrix $A_G$ of $G$. Then, the matrix $A_G^n$ (i.e., the ...

1

In contrast, Non-Euclidean geometries are not usually called by the name of their discoverers. It is fairly standard to refer to hyperbolic geometry as Lobachevskian and I have also seen Bolyai's name included, too. While I didn't know this until I did a bit of web searching, elliptic geometry is sometimes called Riemannian, although one can see how the ...

1

The diagram below uses a slightly different labeling ($A$, $B$, and $C$ are opposite $L$, $M$, and $N$ respectively). Here's a computational justification of the intersection method. (It's possible there are simpler arguments.) Let $(x, y, z)$ denote spatial Cartesian coordinates. Without loss of generality, the triangle lies in the plane with equation $x + ... 1 Because the vertex of a parabola is the point$(\bar{x},\bar{y})$such that$\bar{y}$is maximum (if$a < 0$) or minimum (if$a > 0$), and these situations both happen for$\bar{x} = h$, which ultimately leads to$\bar{y} = k$. You can also think of the symmetry of the parabola in the$y$-axis. Meaning, find the roots of$a(x-h)^2+k$and take their ... 1 First, to answer the question about drawing the propositions: Every "given" that you start with at the outset of a proposition is supposed to be guaranteed by the five postulates. The first three postulates can be oversimplified to "moves you can do with a straightedge and a compass" as you have probably noticed. So that means you will always be able to ... 1 Pick an origin for$\Bbb E^n$, and call the resulting vector space$\Bbb R^n$. To save notation, let's put the origin in$S$, so that$S$now becomes a (vector) subspace. Write$\Bbb R^n=S\oplus S^\perp$. Now, identifying points$x\in\Bbb E^n$with the corresponding vectors in$\Bbb R^n$, write$x=x_1+x_2$, where$x_1\in S$and$x_2\in S^\perp\$. Then ...

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