# Tag Info

24

This is a natural question; the short answer is (1) yes, and (2) that this can be an instructive and powerful way to understand particular groups. In fact, this perspective is so natural, that modern students are sometimes surprised that groups were not invented for this purpose. (Rather, Galois introduced them to study what are now called Galois groups, ...

6

Your solution posted in the given link is so nice that it deserves to come alive here. Consider a four-dimensional ball of radius $R$: $$B:=\bigl\{(x,y,u,v)\>\bigm|\>x^2+y^2+u^2+v^2\leq R^2\bigr\}\ ,$$ and at the same time a four-dimensional dicone $$C:=\bigl\{(x,y,u,v)\>\bigm|\>u^2+v^2\leq x^2+y^2\leq R^2\bigr\}\ .$$ Due to symmetry the dicone ...

4

Your expression is indeed wrong, since $\forall l\forall P:P\notin l$ means that "no point is on a line". What you want is in fact: $$\forall l\forall P\big(P\notin l\to\exists !m\left(P\in m\land m||l\right)\big)$$ which translates as: "for any line $l$, if a point is not on this line, then there is a unique line $m$ to which this point belongs and which is ...

3

We may assume without loss of generality that $CA\geq CB$. Then the triangle $ABC$ lies inside the circle $\Gamma$ with centre $C$ and radius $CA$. Assuming $CD>CA$, $D$ lies outside $\Gamma$, but that leads to a contradiction, since a circle is a convex set and the segment $AB$ lies inside $\Gamma$. Another approach: assume that the projection of $C$ on ...

2

You are on the right track. By assuming that all the possible triples are associated with an obtuse triangle, it follows that: $$x_3^2 \geq x_1^2+x_2^2,\quad x_4^2\geq x_3^2+x_2^2 \geq x_1^2+2x_2^2,\quad x_5^2\geq x_4^2+x_3^2 \geq 2x_1^2+3x_2^2$$ but the last inequality contradicts: $$x_5^2 \leq (x_1+x_2)^2 = x_1^2+2x_1 x_2+x_2^2 \leq x_1^2+3x_2^2.$$

2

Three-dimensional case (n=3) The Wikipedia article on area and volume element states that the area element of the unit sphere is $\sin\phi\,\mathrm d\phi\,\mathrm d\theta$ where I use $\phi$ to denote inclination and $\theta$ to denote azimuth. If you consider your fixed vector as the zenith direction, then my $\phi$ will be the same as yours. To turn ...

2

There are some simple cases. For $r>1$, the hypercube is big enough that the hypersphere will only intersect half its faces. The hypercube will cut out $2^{-n}$ of the hypersphere. So in the plane, you'd cut out a quadrter of the circle, in 3d you'd get one quarter of a sphere, and so on. Combine that with the volume of the $n$-ball and you get ...

2

The logical negation of the parallel postulate is: There exists a line L and a point P not on L such that $|\{$lines parallel to L passing through P$\}| \neq 1$. However, the space we are trying to describe should be sufficiently homogenous that this is equivalent to: Given a line L and a point P not on L, $|\{$lines parallel to L passing through P$\}| ... 2$CE/CD=CF/CB=2\angle ECF=\angle DCB$These triangles are similar. Including BD parallel to FE =>$BD/FE=2$=>$BD=20$Diagonals in a rhombus are always perpendicular.$X=FE \cap AC$, but FE perpendicular to the AC. => AXF right-angled triangle.$AF=13FX=10/2=5$from the similarity. =>$AX=(169-25)^{1/2}=12=3/4*AC$from the similarity. => ... 2 I hope those hints will help you. 2 This theorem is known as Urquhart's theorem (or at least half of it). There is an elementary synthetic solution, which you can find at Cut The Knot. 2 Here is a somewhat efficient approach: Build a matrix$P$of distances,$P_{ij} = P_{ji} = d(p_i, p_j)$Keep an array$d_i = \min_{q\in Q} d(p_i, q)$and an array of lists$Q_i = \{q\in Q \mid d(p_i,q) = d_i\}$. We'll build these on-the-fly For$i = 1$perform the normal search. For each$i \ge 2$: Find$\bar d = \min_{j = 1}^{i-1} P_{ij} + d_j$and ... 2 Draw a line from the centre of the larger circle to the left-most corner of the shaded region. This has length$2r$. We then have an obtuse angled triangle with largest angle$\frac {2\pi}{3}$, due to symmetry. Let$\theta$be the angle between the radii. Using the sine rule, we have $$\sin\frac {2\pi}{3}=2\sin(\frac{\pi}{3}-\theta)$$ Expanding and ... 2 WLOG let$r=1$. Let us use a coordinate system centered on the large circle. Then the oblique line on the right has equation$x=\sqrt3(y-1)$and meets the circle$x^2+y^2=4$. Solving the quadratic equation, the intersection point is$(\dfrac{\sqrt{39}-\sqrt3}4,\dfrac{\sqrt{13}+3}4)$. Hence, the aperture angle of the large sector ... 2 When two lines intersect, you have that opposite angles are equal, therefore if one is a right angle then the opposite is right too. We also know that two adjacent angles (of this intersection) add up to$180$degrees, so if one is right, its adjacent one should also be right as to add up to$180$. Finally, the opposite of this one is equally right. 2 From the Pentagon article on Wikipedia: 2 Lemma: Let$x,y,z>0$. Then there exists a triangle with sides$x,y,z$if and only if$2x^2y^2+2y^2z^2+2z^2x^2>x^4+y^4+z^4$. Proof of lemma: Observe that$2x^2y^2+2y^2z^2+2z^2x^2-x^4+y^4+z^4=(x+y+z)(-x+y+z)(x-y+z)(x+y-z)$. If there exists a triangle with sides$x,y,z$, then each factor is positive, so their product is positive. Conversely, if there is ... 2 One way to interpret your "equality" is from the point of view of convex analysis. But first talk linear algebra: Let$\vec{v}$be the vector$y - x$and let$\vec{w}$be the vector$z - y$, we have by definition that the vector$\vec{u} = z - x$can be written as $$\vec{u} = \vec{v} + \vec{w}$$ Your equality can be re-written as the statement $$... 2 I believe you understood the proof. The proof is a general case where AC greater than AB. You can place the sides AC and AB anywhere in the picture. For example, switching the letters A and C in the picture, you get the proof of another pair. And notice the proof does not depend on the position of the letters. 2 This GIF (jiff) is very instructive. 1 \newcommand{\Reals}{\mathbf{R}}Even in one dimension, a "sphere" has equation x^{2} = r^{2}, i.e., x = \pm r. If I understand, your friend's question is this: Let n be a positive integer, x_{1}, ..., x_{n} Cartesian coordinates, and define the sphere of suidar r to be the set of points in \Reals^{n} satisfying$$ x_{1}^{2} + \dots + ... 1 No. This isn't even true for linear transformations; take$A$to be any nontrivial rotation about the origin,$B=I$, and$v=0$. Then$ABv=v$but$AB\neq I$. 1 Let$M$be a set and$d\colon M^2\to ]-\infty,\infty[$be a symmetric map which satisfies the “goat axiom”: $$d(x,y)\leq d(x,z)+d(y,z),$$ then$d$is a metric, i. e., it can be shown that$d$is non-negative and satisfies the triangle equation$d(x,y)\leq d(x,z)+d(z,y)$. Why is it called goat axiom? Two goats, located at$x$and$y$are tied to a peg ... 1 i will give you some hints. (a) can you find the slope of$DF?$(b) let$P$be the foot of the perpendicular from$E$to$DF.$what is the relation between the slopes of$EP$and$DF?$(c) you know the slope of$EP$and the point$E=(7,3).$can you write the equation of$EP?$(d) can you also write the equation of the$DF?$(e) can you solve the two ... 1 To get you started: The slope between two points$(x_1,y_1)$and$(x_2,y_2)$is$\dfrac{y_2-y_1}{x_2-x_1}$. Use this to get the slope of$DF$. The altitude from$E$is a line through$E$perpendicular to the opposite side of the triangle (i.e.,$DF$). Remember that if a given line$\ell$has slope$m$, then a line perpendicular to$\ell$has slope ... 1 This just comes down to computing an isosceles triangle because $$\|\vec{P}_1 - \vec{C}\| = \|\vec{P}_2 - \vec{C}\| = d$$ and the angles at$\vec{P}_1$and$\vec{P}_2$, let's call them$\beta$, are equal.$\|\vec{P}_1 - \vec{P}_2\| =: a$is known. Use basic trigonometry and you're done. From Pythagoras we get the triangle's height from $$h = \sqrt{d^2 - ... 1 A rotation by an angle -\theta (i.e. clockwise) around a point C can be interpreted as a translation from C to the origin, followed by a rotation of angle -\theta around said origin, and a final translation from the origin back to C. Thus:$$R^{-\theta}_C=T^{-1}\circ R^{-\theta}_O\circ T$$... 1 Here's a cleaner version of my attempt at an answer. It's still not complete but it's free of clutter and also identifies the circle M is supposed to lie on. I also tried a solution based entirely on analytical geometry and I got very far but it's very messy. The picture below is a screen shot of an animation based on my analytical treatment. Too bad I ... 1 Let's label the shortest path red and then flatten the shape: So we need to minimize the path:$$\min\limits_{\theta}\left(\sqrt{h^2+(2r\theta)^2}+2r\cos\theta\right) = 10\,\hbox{cm} \cdot\min\limits_{\theta}\left(\sqrt{\frac{1}{4}+\theta^2}+\cos\theta\right)=15\,\hbox{cm},$$according to wolframalpha. 1 If I understand the question properly, open up the cylindar to form a rectangle$5{\rm{ by 20}}\pi $A is one corner of the rectangle and B is half way along the${\rm{20}}\pi $side. Using pythagoras gives a distance around the cylindar of$\sqrt {100\mathop \pi \nolimits^2 + \mathop 5\nolimits^2 } \$

Only top voted, non community-wiki answers of a minimum length are eligible