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0

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.


4

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 ...


3

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


3

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.


5

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 ...


15

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 ...


2

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 $x^3-x$


0

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 ...


3

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 $$ 2=(3a^2-1)(-2)-2a^3. $$


0

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. ...


3

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 ...


0

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, ...


2

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) = ...


27

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 ...


-3

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


0

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 ...


0

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


0

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 ...


1

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 ...


0

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.


0

Put the first point at the top of the circle. Randomly select the second point uniformly from the circle; without loss of generality, assume the point falls on the right half. (If it's on the left, the arguments that follow can just be mirror-reversed.) Let the angle between the two points (as measured from the circle's center) be $x$ radians; clearly, $0 ...


1

The distance you want is the shortest distance which is perpendicular to both lines. Take any point on the two lines in terms of any parameter you want to assign say $k_1$ and $k_2$ suppose points are $P_1:(f_1(k_1),f_2(k_1),f_3(k_1)$ on $L_1$ and $P_2:(f_4(k_2),f_5(k_2),f_6(k_2)$ on $L_2$ . The direction cosine joining the two point, $P_2-P_1$ and if you ...


2

For any points $p$ and $q$, $$ \{x\in\mathbb R^2 \mid \|x-p\|\le\|x-q\| \} $$ is a halfplane (or the whole plane if $p=q$), so your set is an intersection of halfplanes, hence convex.


0

I'd recommend simplifying as much as possible at each stage: $$ r = (\sin{t},\cos{t}+\log{(\tan{\tfrac{1}{2}}t)}) $$ then $$ \dot{r} = (\cos{t},-\sin{t}+\csc{t}) = \left(\cos{t},\frac{1-\sin^2{t}}{\sin{t}} \right) = (\cos{t},\cos{t}\cot{t}). $$ Finally, $$ \ddot{r} = \left( -\sin{t}, -\cos{t}-\cot{t}\csc{t} \right) $$ and so $$ \lvert \dot{r} \rvert^2 = ...


0

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\setA}{\mathcal{A}}$If the points $(x_{j})_{j=1}^{N}$ are linearly independent as vectors in $\Reals^{n}$, the set $\setA$ is the parallelipiped[1] centered at the origin, with a vertex at $-\sum_{j} x_{j}$ and edges $(2x_{j})_{j=1}^{N}$.[2] If the points are linearly dependent, $\setA$ is a projection of a ...


1

In the denominator of your full expression for $\kappa$ there should be a $-2$ rather than $-1$


1

Zonotope. (I have nothing more to say, but say more to satisfy the computer.)


1

Hint: In a right triangle, knowing two sides is enough to determine the third side. For the ladder to reach the window, the third side, which represents the distance from the ground to the point of contact of the ladder with the house, must be at least the height of the window above ground. You will use the Pythagorean Theorem to find the third side. Note ...


1

Another point of view (which is the theme of John Stillwell's book "Four pillars of modern geometry"), is a synthetic derivation that starts with the upper half plane equipped with its group of rigid motions: namely, the group generated by the fractional linear action by $PSL_2(\mathbb{R})$ together with reflection across the upper $y$-axis. From this, ...


8

Both the foci of a circle coincide and thus, its eccentricity is zero. So yes, it is an ellipse. It is like that all squares are rectangles but all rectangles are not squares. See this link- [the most intuitive link ever seen!] http://www.mathsisfun.com/geometry/eccentricity.html


-2

This is just an imaginary point. In practice, you can go on following the lines upto whatever distance you wish, I bet your generations will never find that intersection point. There is no point like that. In books they put it to relate it to algebra which can be expressed mathematically only. But then the question is why do we say that ? We say that because ...


0

Why do lines in the poincare model meet the infinite edge at right angles? Interesting question you can look at it from many different points: Axiomatic Trough very two points there is one line and one line only, and when the limitation was not in place then trough two points there could be more than one line. (and making the whole model a geometrical ...


2

Required to prove that the acute angles made by each foci to the tangent are equal


9

Yes. A circle is a special case of an ellipse. The equation of an ellipse is: $$\left(\frac xa \right)^2 + \left(\frac yb \right)^2=1$$ When $a=b=1$, this gives the equation of the unit circle: $x^2+y^2=1$. In general, if $a=b=r$, you get the equation of a circle with radius $r$: $$x^2+y^2=r^2$$


1

In polar coordinates: $$ \vec{OM} = r\hat e_r\\ \vec v = r' \hat e_r + r \omega \hat e_\theta \simeq r \omega \hat e_\theta \\ \vec a = [r'' - r\omega^2] \hat e_r + [2r' \theta' + r\omega'] \hat e_\theta \simeq - r\omega^2 \hat e_r + r\omega' \hat e_\theta \\ \implies r = \frac{\|\vec v\|^2} {\| \vec a \wedge \frac{\vec v}{\|\vec v\|} \|} $$ Write ...


-2

a triangle has $180^\circ$ a staight line has $180^\circ$ opposite angles are equal


0

Note that for the point $p$ to lie within the spherical triangle bound by $\{ p_1, p_2, p_3 \}$, the intersection of the ray from the origin to $p$ (basically the line joining $(0,0,0)$ and $p$) with the 3D-plane defined by $\{ p_1, p_2, p_3 \}$ lies inside the planar triangle $\{ p_1, p_2, p_3 \}$. This can be formulated with the unknown quantities ...


1

Let one point be fixed and let circle have radius $1$. Now find posibility for other two points. First integrate all possibilities: $$I_1=\int_0^{2\pi}\int_0^{2\pi}dbda=4\pi^2$$ Then integrate required case: $$I_2=\int_0^{\pi}\int_a^{2\pi}dbda+\int_{\pi}^{2\pi}\int_0^{a}dbda=3\pi^2$$ So, your probability is $\frac {I_2} {I_1}=\frac34$. Explanation: let ...


1

$dr$ is the infinitesimal change in the radial direction while $r d\theta$ is the corresponding change in the perpendicular direction. The arc length is therefore: $$ ds = \sqrt{dr^2 + (rd\theta)^2} = \\ \sqrt{1 + \left(r\frac{d\theta}{dr}\right)^2}dr $$ Also, $tan\beta$ is the ratio of tangential to the radial displacement i.e. $$ tan\beta = ...


1

First of all, note that $EA=AB$. Take a point $G$ on $DB$ such that $\angle{GAB}=20^\circ$. Since $\angle{ABG}=\angle{AGB}=80^\circ$, one has $AB=AG$. So, since $\triangle EAG$ is an equilateral triangle, one has $AG=GE$. Since $\angle{GAD}=\angle{GDA}=40^\circ$, one has $AG=GD$. Hence, one has $GE=GD$. So, since ...



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