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As it turns out, several enterprising gentlemen discussing the original problem found a provably optimal $\mathcal{O}\left( n^2\right)$ algorithm, which can be generalized to the $p$-dimensional case, $p \geq 3$ via Dilworth's theorem. Hence this resolves question 2. The question about whether this box-grouping problem or others similar to it have a formal ...


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The spin group is the double cover of the orthogonal group. In dimension three, the spinors are $SU(2)$, which is the double cover of $SO(3)$. It can be identified with the unit quaternions, and is homeomorphic to a three-sphere. ($SO(3)$ is homeomorphic to the three-dimensional real projective space.)


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Yes, quaternions are related to spinous in three dimensions. You might first try the subsection of the spinor wiki talking about quaternions where it mentions Thus the (real) spinors in three-dimensions are quaternions, and the action of an even-graded element on a spinor is given by ordinary quaternionic multiplication. See also ...


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Just join middle point of AB to O ;it is now easier to think out the answer than ask this question.


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Edited your sketch. When two line segments cut at B, use circle segments property of constant products: $ BA . BD = BC. BE ; 2 ( 2 R -2) = 4^2 $ so $ R = 5. $


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Gradient The direction of the cylinder is the direction where the value of the left hand side does not change. So $(a,b,c)$ is the direction if $$\forall x,y,z\in\mathbb R:f(x,y,z)=f(x+a,y+b,z+c)$$ where $f(x,y,z)$ denotes the left hand side of your equation. The change of $f$ in a given direction can also be described using the gradient $\nabla f$. What ...


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This is supposed to be a comment but I would like to post a picture. For any $m \ge 3$, we can put $m+2$ vertices on the unit sphere $$( 0, 0, \pm 1) \quad\text{ and }\quad \left( \cos\frac{2\pi k}{m}, \sin\frac{2\pi k}{m}, 0 \right) \quad\text{ for }\quad 0 \le k < m$$ Their convex hull will be a $m$-gonal bipyramid which appear below. Up to my ...


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$A+t(B-A)$, $B$ is at $t=1$............................


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Let $p=(x_0,y_0)$ use double angle formula $g(\theta ) = \cos^2 (\theta )f_{xx} (p) + 2 \cos(\theta )\sin(\theta ) f_{xy} (p) + \sin^2 (\theta ) f_{yy} (p )= \frac{1}{2}(f_{xx} (p )+f_{yy} (p))+\sqrt{\frac{1}{4}(f_{xx} (p)-f_{yy} (p))^2+f_{xy}(p)^2} \cos(2\theta+\alpha)$ for some constant $\alpha$. So the product of minimum and maximum value is ...


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Let $\overline{CC^\prime}$ be a diameter. By Power of a Point, $AB \cdot BD = CB \cdot BC^\prime = 4 \cdot 4 = 2 \cdot BC^\prime$. Then $BC^\prime = 8$, $CC^\prime = 10$, and $CO = 5$.


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$PH$ must split the trapezoid into two trapezoids. For the sake of convenience, assume $x$ is the bottom side of the trapezoid and $4$ is the top side. Of these two trapezoids, label the height of the top trapezoid as $A$ and the bottom as $a$. Since they have the same area, we have from the trapezoid area formula the equality $$5A=\frac{(6+x)a}2$$ Since we ...


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Approximate image using manual tuning Regarding your question of how the correct drawing would look like: This was created using Cinderella with manual tuning of the configuration till it met the constraints you mentioned. Exact coordinates from lengthy computation You could compute coordinates for the corners, and then think of ways to construct the ...


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If $u,v,w$ are the vector fields $\partial_i,\partial_j,\partial_k$ then the term $\nabla_{[\partial_i,\partial_j]}\partial_k$ vanishes. This is because $[\partial_i,\partial_j]=0.$ However, this doesn't hold for arbitrary vectors. Note that in the Riemann curvature tensor there are involved second derivatives of a vector. So: $$\nabla_u (\nabla_v ...


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I'll combine three parts. Let's start with the fractional part function, and adapt that so it maps $\mathbb R$ to $[-1,1)$: $$f_1:t\mapsto 2(t-\lfloor t\rfloor)-1$$ This ensures a period of $\tau=1$. The next part is a rational parametrization of the semicircle, which maps $[-1,1)$ to the right half of a circle with radius $2$ and center $(-1,0)$: ...


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Because $D$ lies on the circle, $OD^2=OB^2+BD^2$. Also because $OD(=OC)=OB+BC$, $OB=OD-BC$. Substituting $OB$ in the first equation yields $ $ $ $ $ $ $ $ $OD^2=(OD-BC)^2+BD^2$ $≡$ $OD^2= OD^2-2\centerdot OD\centerdot BC+ BC^2+BD^2$ $≡$ $2\centerdot OD\centerdot BC =BC^2+BD^2$ $≡$ $OD = (BC^2 + BD^2)/(2\centerdot BC)$ For the example at hand, this ...


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If you define a parallelogram to have two pairs of parallel sides then a square will satisfy this definition thus all squares are parallelograms. But this does not mean all parallelograms are squares.


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Hint: Suppose $OB=x$, then $OA=OC=x+2$. Now apply Pythagoras theorem in triangle $AOB$


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Let $CO=r$ then we have $r^2=(r-2)^2+4^2$. Solve for $r=5$


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Let $P=(-70,3)$ and $Q=(88,81)$, and let $\vec{a}=\vec{PQ}=\langle158,78\rangle$. Using the coordinates of $P$, we obtain $x=-70+158t, y=3+78t$ as parametric equations for the line.


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Take two of those points, $A$ and $B$, and set $f(t)=A+t(B-A)$. Then split into equations for x and y.


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I thin you overworked here. The condition is $$|AP|^2=9|BP|^2\iff (x+3a)^2+y^2=9\left[(x-a)^2+y^2\right]\iff$$ $$\iff 8x^2-24ax+8y^2=0\iff \left(x-\frac{3a}2\right)^2+y^2=\frac{9a^2}4$$ and we get that the points$\;P=(x,y)\;$ are the locus of circle with center $\;\left(\frac{3a}2\,,\,0\right)\;$ and radius $\;\frac{3a}2\;$ , and since radius = absolute ...


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You know the angle $C$ and $A$ are right and $BD$ must be perpendicular to $AC$(since the length of BC and AB are the same). Suppose the intersection point is $M$. Hence you can $AM$ by using BCM is similar to BDC.


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Since the line is perpendicular to the plane, so is any nonzero vector parallel to the line, including, the vector from $(1, -1, 0)$ to $(2, 1, 0)$, namely, $${\bf n} := ( 2 - 1, 1 - (-1), 0 - 0 ) = ( 1, 2, 0 ).$$ Now, by definition any point $\bf x$ is in the plane if the vector ${\bf x} - {\bf x}_0$ from ${\bf x}_0 := (0, 1, 1)$ to $\bf{x} = (x, y, z)$ ...


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A direction vector of the line is $$\begin{pmatrix}2-1\\1--1\\0-0\end{pmatrix}=\begin{pmatrix}1\\2\\0\end{pmatrix}$$ If the plane is perpendicular to the line, the normal vector of the plane is equal to the direction vector of the line (convince yourself of this). A plane equation with normal vector $(a,b,c)$ and passing through the point $(x_0,y_0,z_0)$ ...


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You don't need the equation of the line, only its direction vector, for this vector is the normal vector of the plane. $$\vec n=(2,1,0)-(1,-1,0)=(1,2,0)$$ Now, the equation of the plane is $$x+2y=C$$ To find $C$ you only have to subst the point the plane passes through. $$0+2\cdot1=C\implies C=2$$


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MATHEMATICAL ANSWER: Referring to the picture: let $\theta$ be the angle $B\widehat{A}P$, in your case $\theta=\pi/6$. $f(h)$ describes the shape of an edge of the dome as a function of the height, i.e. $f(h)$ is the distance of the edge from the central axis of the dome evaluated at a ceratin height $h$. Let's suppose that you know $f$ and let's suppose ...


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There is nothing wrong in the steps illustrated.


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Hint For an horizontal cylinder, if $L$ is the length, $R$ the radius and $h$ the height of the liquid, the volume occupied by the liquid is given by $$V=L \left(R^2\cos ^{-1}\left(\frac{R-h}{R}\right)-(R-h) \sqrt{2 h R-h^2}\right)$$ So, you need to solve something for $h$.


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Here is an article on the Goedel's lost letters blog, which deals with elementary identities, which cannot be proven using "high school mathematics", for how "high school mathematics" is defined see, http://rjlipton.wordpress.com/2014/07/10/high-school-theorems/ Especially see the HSI section.


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Let $r$ be the radius, $L$ be be the horizontal length of the cylinder and $h_0$ be the height of the level of the fuel above the centre of the cylinder(note that $h_0 + 1.25$ is the required level). After $3 m^3$ fuel has been taken out the volume of the remaining fuel is given by: $$ \left\{\frac{\pi r^2 L}{2}+2\cdot L\int ^{h_0}_0\sqrt{r^2-x^2} \ ...


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If I understand you correctly, you are seeking for formal systems of first order logic in which to formalize statements of Euclidean geometry? There seem to be several ways to do this. Firstly, you might formalize statements of Euclidean geometry inside a FOL set theory like ZFC, taking ${\mathbb R}^2$ as the domain of discourse, and prove them there ...


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Natural deduction is a formal proof calculus, thus we can apply it to any formal theory.


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Area of wall = base × height = 30×10=$300m^2$ $\because 5m^2$ is covered by 1 litre of paint $\therefore 300m^2$ is covered by $\frac 15$ × 300 = 60 litres of paint. $\therefore$ cost of painting = 60 × 20 = $\$ 1200$.


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Hint : A parallelogram can be transformed into a rectangle, so the problem boils down to find the surface of a rectangle. From http://rioranchomathcamp.com/PerimeterAreaVolume.asp.


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I used Cinderella to apply a projective transformation which maps $B$ to $B'=(0,0)$, $G$ to $G'=(64,0)$, $A$ to $A'=(0,32)$ and $E$ to $E'=(64,32)$. That map applied to $P$, the midpoint of your ball, is a point approximately at $P'\approx(31.58,13.95)$ so the coordinates of $P$ in 3d should be approximately $(0,13.95,31.58)$. From the picture you can see ...


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It is because in any neighborhood of a boundary point none of the orthogonal projections to the coordinate plane give a parametrization.


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First off, let's clarify the relation to quaternions. If you have \begin{align*} x &= x_1 + x_2\mathbb i + x_3\mathbb j + x_4\mathbb k \\ y &= y_1 + y_2\mathbb i + y_3\mathbb j + y_4\mathbb k \\ z &= z_1 + z_2\mathbb i + z_3\mathbb j + z_4\mathbb k \end{align*} with all the coefficients on the right chosen real, and furthermore have $z=x\cdot ...


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Since this is most definitely a homework question, I shall give you necessary hints and theorems to prove it yourself. The answer is that area of A'B'C = $\frac {1}{4}$ Area of triangle ABC. Theorem: Two triangles are similar if two of their corresponding sides are proportional and the angle between the two sides is equal. Theorem: The ratio of areas ...


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Wrong question At first I missed the “on a line” part of your question, and assumed that $P$ would be on a circle of radius $3$ around the center of the stated circle. The computation below originally refers to that. This problem has perfect rotational symmetry: you have that circle of radius $1$ and another of radius $3$ and $P$ can be at any position on ...


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$CA'/CA=CB'/CB=BA'/BA$ 1/2=1/2=1/2 so Areaof triangle ABC=1/2*2*2=2 Area of triangle A'B'C=1/2*1*1=1/2 SO Area of triangle A'B'C=1/4 *area of triangle ABC


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The simplest way is to use polar coordinates to obtain the centers of $n$ circles equidistant from each others. These coordinates are $$(\rho \cos\frac{2 \pi k}{n}, \rho \sin\frac{2 \pi k}{n})$$ where $\rho$ is the radius of the circles and $k=0,1,2...(n-1)$ is an integer that identifies the $k^{the}$ circle, counting them from that corresponding to $k=0$ ...


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in the 3D-coordinate system (x,y,z): write the line that joins $p_1p_2$ find the point of this line with $z$-coordinate $=0$ (intersection with your plane) done


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Hint. Draw the lines $PA,PB,PC,PD$ and also the perpendiculars from $P$ to the sides of the rectangle. Then you will see eight right-angled triangles. Now use Pythagoras' Theorem (more than once).


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As you already pointed out, sequences of the type $\left\{\left(-\dfrac{1} {n},y\right)\right\}$ are not equivalent to $\left\{\left(\dfrac{1} {n},y\right)\right\}$, for $|y|\leq1$, so each one will converge to a different point in the complete space. If we call this points $(0^-,y)$ and $(0^+,y)$ their distances will be given by $d=2(1-|y|)$. So the space ...


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The main trick here is $$(x+y) \otimes (x+y) - x \otimes x - y \otimes y = x \otimes y + y \otimes x $$ I bet you're making things much harder by trying to do things in terms of a basis, rather than using tensor algebra. Also, you're probably contributing to your confusion by identifying elements of $V \otimes V$ with as bilinear forms on $V$: it would be ...


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To answer it just let me put a caveat: what is General Relativity may be open to interpretation as to the extend you would like to add astrophysics and cosmology to the definition. This more "applied" fields depend a lot on other parts of physics and would carry whatever mathematical techniques this areas involve. Also it would depend if you want to consider ...


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This is a variant on AsdrubalBeltran's answer, starting from the equation $$-2+\sqrt{3}=\tan(165^\circ)=\tan(3\cdot55^\circ)=\frac{3\tan(55^\circ)-\tan^3(55^\circ)}{1-3\tan^2(55^\circ)}$$ This implies that $\tan(55^\circ)$ is a root of the cubic polynomial $$P(t)=t^3+(6-3\sqrt3)t^2-3t+(\sqrt3-2)$$ Note that $$P(0)=\sqrt3-2\lt0$$ and ...


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I found a solution: Note that : $\tan(3\cdot x)=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}$ Then $$-2+\sqrt{3}=\tan(165^\circ)=\tan(3\cdot55^\circ)=\frac{3\tan(55^\circ)-\tan^3(55^\circ)}{1-3\tan^2(55^\circ)}$$ Now how: $$-\frac{18}{73}=\tan(3\cdot\tan^{-1}(1.5))=\frac{3\cdot1.5-(1.5)^3}{1-3(1.5)^2}$$ but it is known that $\tan{3x}$ is increasign, if ...


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No, it cannot. Consider the line $x-y=0$ (i.e. $a=1, b=-1, c=0$). Consider the two points on the line : $A(0,0)$ and $B(1,1)$. Consider now the line $r$ : $y=0$ (the $x$ axis) and on it the point $A'(0,0)$. Then, you cannot find on $r$ a point $B'$ such that $A'B' \equiv AB$ because the lenght of this segment is $\sqrt 2$ and the point with coordinates ...


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OB=OA=OC=radius of the circle centered at O AB=AO=AC=radius of the circle centered at A You have that AO=OA so all the above line segments have equal lengths.



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