# Tag Info

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This quantity is much studied in high-dimensional convex geometry; an excellent starting point is Keith Ball's An Elementary Introduction to Modern Convex Geometry (MSRI, 1997), especially Lecture 2. See also this question.

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You can use Ptolemy's theorem: A quadrilateral is inscribable in a circle if and only if the product of the measures of its diagonals is equal to the sum of the products of the measures of the pairs of opposite sides. In our case, it is obvious from mental diagram that diagonals are $\overline{(9, 6)(1, -2)}$ and $\overline{(0, 0)(4, -4)}$, and ...

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let us call the points $$A = (0,0), B= (1,-2), C = (4, -4), D = (9, 6)$$ you need to vrify that $$2\cos \angle BCD = \frac{BC^2 + CD^2 - BD^2}{BC \cdot CD} = -\frac{AB^2 + AD^2-BD^2}{AB \cdot AD} = -2\cos \angle BAD$$

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It is enough to find two opposite vertices whose angles add to 180 degrees. Make vectors of the sides, and use the dot product to calculate cosines of the vertex angles. The cosines of opposite vertices need to be equal in magnitude, but opposite in sign.

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Use the property that the perpendicular bisectors of two cords on a circle intersect at the centre. A line passing through $(9,6)$ and $(4,-4)$ is $2x-12$. The perpendicular bisector of that segment is thus $-\frac12x+{\frac {17}{4}}$. Likewise, the line passing through $(0,0)$ and $(4,-4)$ is $-x$, and its perpendicular bisector is $x-4$. The intersection ...

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This is a natural question, and can be a very instructive way to understand particular groups. In fact, this perspective is so natural, that to modern students it is somewhat surprising that groups were not invented for this purpose. (Rather, Galois introduced them to study what are now called Galois groups, that is, the groups of automorphisms of splitting ...

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the eqution of line is $$\frac{y-y_0}{x-x_0}=slope$$ $$\frac{x^3-x-2}{x+2}=3x^2-1$$ $$3x^3-x+6x^2-2=x^3-x-2$$ $$2x^3+6x^2=0$$ $$2x^2(x+3)=0$$ if $$x=0$$ or $$x=-3$$ that means there are two lines tangent to the $x^3-x$

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There are two approaches to this: you can find all the tangents that pass through $(-2,2)$, or find all the lines through $(-2,2)$. Both should end up with the same equations. All lines that pass through $(-2,2)$ are of the form $$y-2 = m(x+2).$$ These intersect the curve $y=x^3-x$ when $$x^3-x-2 = m(x+2).$$ For an intersection to be a tangent, the ...

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Given a point on the curve, it has the form $(a,a^3-a)$. And the slope of the tangent at that point is $3a^2-1$. Then, using point-slope form for a line, the tangent line has the form $$y=(3a^2-1)(x-a)+a^3-a.$$ Expanding and simplifying, this becomes $$y=(3a^2-1)x-2a^3.$$ If $(-2,2)$ is on this line, we must have $$2=(3a^2-1)(-2)-2a^3.$$

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Multidimensional spaces occur naturally all around us. Take the possible positions of your arm, for example. You can rotate your shoulder with two degrees of freedom; You can bend your elbow with one degree of freedom; You can rotate your wrist with two degrees of freedom That's a total of five degrees of freedom for the possible positions of your arm. ...

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There have been people who reportedly can visualize things in four dimensions as easily as other people can in three. It's rare, however. Moreover, visualizing four dimensions may not help much when you want to solve a problem in five dimensions or more. So as Henning Makholm's answer states, to do anything really useful in higher dimensions you need a ...

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Dimension usually is just the number of 'components' of some piece of information. 3 dimensions are just nice for describing a position in (Euclidean) space, but you definitely need 4 dimensions if you want to include the time also. Now you are in the room. A while later you are not. Your position has changed over time, so if we want to describe your path, ...

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Let's say that $AB=3x$, $CD=2y$ and $BC=h$. Then, $$3xh = 2yh = 70,$$and $$x = \frac{2y}{3}.$$ The triangles $AFH$ and $HEC$ are similar and have area: $$S_{AFH} = \frac{x h_1}{2}, S_{HEC} = \frac{y h_2}{2}$$ where $h_1$ and $h_2$ are the heights of the 2 triangles, with $h_1+h_2 = h$ and $h_1 = \frac{2h_2}{3}$. Then: $$h_1 = \frac{2h}{5}, h_2 = ... 1 There's a "straightforward" vector solution: Let AB=b, AD=d the basis vectors and [x\times y] be the cross product, since we're given |[AB\times AD]|= 70. Vectors AF=1/3AB=1/3b, AG=2/3AB=2/3b, AE=AD+1/2DE=AD+1/2AB=d+1/2b. X lies on the line YZ iff AX=t\cdot AY + (1-t)\cdot AZ, where t is a real. So, consider AH=(1-u)\cdot 0+u\cdot ... 1 The morphism \varphi_C is best seen as a morphism \varphi_C:C\rightarrow(\mathbb{P}^2)^* from the curve C\subset \mathbb P^2 to the dual projective space (\mathbb{P}^2)^*, whose typical point Q=(a:b:c)\in(\mathbb{P}^2)^* corresponds to the line l_Q\subset \mathbb P^2 of equation ax+by+cz=0. If C is a line ax+by+cz=0, then the morphism ... 0 One way of doing it: The chords are the third side of triangles whose other two sides are radii of the circle. For the two chords you know the length of, you thus have triangles where you know the length of all three sides, using e.g. the law of cosines you can find the angle they span the center of the circle. Subtracting those angles from 2\pi gives you ... 0 Hint: use law of sines. R here is the circumradius of the given triangle. ^^ \dfrac{a}{\sin A} = 2R"  similarly with other two sides. Sorry, I am on mobile its hard to write. 1 David Foster Wallace in everything and more, a compact history of ∞: There is something I "know," which is that spatial dimensions beyond the Big 3 exist. I can even construct a tesseract or a hypercube out of cardboard. A weird sort of cube-within-a-cube, a tesseract is a 3D projection of a 4D object in the same way that "" is a 2D ... 1 Let the centres of the two circles be (x_1, y_1) and (x_2, y_2), where either x_1 < x_2 or x_1 = x_2 and y_1 < y_2, and radii r_1, r_2. Suppose x_1 = x_2. In this case, the blue line segment is vertical, and its endpoints can easily be seen to be (x_1, y_1 + r_1) and (x_2, y_2 - r_2) = (x_1, y_2 - r_2). Now suppose x_1 < x_2. ... 0 The unit sphere in \mathbb{R}^n is denoted by \mathcal{S}^{n-1}. Its surface area is given by S_{n-1} on Wikipedia, but let's call it A_n here, because, well, we're already using S a lot, and the n-1 is confusing:$$A_n = 2\pi^{n/2} / \Gamma(n/2).$$Now let's look at a different set than yours:$$C(\epsilon) = ...

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The thing is you're not supposed to "wrap your mind around" higher-dimensional shapes. Instead, what happens is that we take a formalism that is made to describe 3-dimensional shapes (which we can understand more or less intuitively), and then we just see what happens when we replace all of the "$3$"s in that theory with "$4$" or "$5$" or more. The outcome ...

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This man can ! https://www.youtube.com/watch?v=M9sbdrPVfOQ :) But I never understand I think we have to have a brain in 4D to understand. You can also see that : A conference really interesting Here And a short video Here

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An alternative way of doing this problem is to consider the reflection of the quadrilateral $ABCD$ in the line $AD$ so that the combined figure is a five-sided polygon (only 5 sides since $BA$ is perpendicular to $DA$). Now the pentagon has one angle of $108$ but now all sides equal $x$. Therefore it is a regular pentagon and therefore $\theta=108$. The ...

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Follow the steps,same as above. The only mistake is in the calculation of angle CBF . Its 18 degrees. So in the final step angle BCD is 108 degrees

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It means that if $X$, $Y$, $Z$ are any three of $B,C,E,F$ then $\angle XYZ$ has measure (in radians) of the shape $r\pi$ for some rational number $r$. Equivalently, since $\pi$ radians is $180^\circ$, it means that every angle is a rational number of degrees. Thus angles like $\frac{99}{7}$ degrees or $17$ degrees are OK, but (for example) an angle of ...

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draw a perpendicular $CE$from $C$ to $AD$ so that the foot $E$ is on $AD.$ draw another perpendicular $BF$ from $B$ to $CE.$ we will compute $EC$ in two ways: $$EC = x \sin 54^\circ = CF + FE = CF + AB = x\sin\angle CBF + \frac x 2$$ therefore $$\sin \angle CBF=\frac{2\sin 54^\circ - 1}{2} \to \angle CBF = 53^\circ$$ therefore $$\angle BCD = ... 0 Parallel mirrors make this essentially a one-dimensional problem, since for any point, all its mirror images will be on a line through that point and perpendicular to all the mirrors. Two mirror reflections combine to a translation by a distance which is twice the distance between the mirrors. Conversely, the group of transformations generated by these two ... 0 The most common way to calculate the distance not using the cross product is: Let \mathbf{r}_1=\mathbf{A}+x\mathbf{l}_1, \mathbf{r}_2=\mathbf{B}+y\mathbf{l}_2 the lines and (\mathbf{a},\mathbf{b}) the inner(or the dot) product of \mathbf{a},\mathbf{b}. (a) We find 2 points \mathbf{r}_1, \mathbf{r}_2 on the lines such, that ... 1 choose the coordinate system with the origin at G. let a, b, -a-b be th e complex numbers representing the vertices A, B, and C. then we have$$\begin{align}|AB|^2 + |BC|^2 + |CA|^2 &= (b-a)(\bar b - \bar a)+(2b+a)(2\bar b +\bar a)+(2a+b)(2\bar a + \bar b)\\ &= b\bar b +a\bar a-a\bar b -b\bar a+4b\bar b + a \bar a+ 2a\bar b+2b\bar a + 4a\bar ...

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Hint: we have $AA'=\frac{1}{2}\sqrt{(b^2+c^2)-a^2}$ and we have $\frac{2}{3}AA'=\frac{1}{3}\sqrt{2(b^2+c^2)-a^2}$ plugging the other formulas in the right side of your equation we get $$3\left(\frac{1}{9}(2(b^2+c^2)-a^2)+\frac{1}{9}(2(a^2+c^2)-b^2)+\frac{1}{9}(2(a^2+b^2)-c^2)\right)=a^2+b^2+c^2$$ after some algebra.

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