# Tag Info

0

Compare the above drawings and determine whose answer is more correct.

0

Well, once we have enough informaton we can try to answer, but so far, as far as I can tell, we don't have enough measurements to answer the question. There's nothing specified that can constrain the BE/AF lengths.

0

From the Wikipedia article on Thales' theorem: If you know the hypotenuse, you know the base of this diagram; if you also know the area, then you can compute the height (from $A=\frac12 bh$). Then you can draw a horizontal line at the required height; where it intersects the circle is the vertex with the right angle.

0

The answer is yes. You have to consider the divisor $D = (2g + 2)q$. Since $deg(D) = 2g + 2 > 2g$ it follows that $D$ is very ample. This means that there is an embedding $\phi: C \to \mathbb{P}^r$ in a projective space $\mathbb{P}^r$ with $r \geq 2$ and a hyperplane $H \subset \mathbb{P}^{r}$ passing exactly through the image $\phi(q)$ which has contact ...

2

For the unit $d$-sphere $S^{d}$ embedded in $\mathbb{R}^{d+1}$, one can parametrize it using $d$ angles $\phi_1,\ldots,\phi_{d}$. $$[0,\pi]^{d-1} \times [-\pi,\pi) \ni (\phi_1,\ldots,\phi_d) \quad\mapsto\quad (x_1,\ldots,x_{d+1}) \in \mathbb{R^{d+1}}$$ where $$x_k = \begin{cases} \cos\phi_1,& k = 1\\ \sin\phi_1\cos\phi_2, & k = 2\\ ... 1 There is a general expression derived in Reflection Formula by HCR for directly calculating the co-ordinates of point of reflection say \color{blue}{(x', y')} of any given point \color{blue}{(x_o, y_o)} in the straight line \color{blue}{y=mx+c} is given as$$\bbox[4pt, border:1px solid blue;]{(x', y')\equiv\left(\frac{(1-m^2)x_o+2m(y_o-c)}{1+m^2}, ...

0

this is really the double angle formula in disguise. the reflection keep the length $1$ of the point $(1,0)$ to $(0,0)$ fixed. therefore the image of $(1,0)$ on the mirror $y = mx$ satisfies the constraint $$x^2 + y^2 = 1. \tag 1$$ now the midpoint $(\frac{1+x}2,\frac y2)$ and twice the midpoint $(x+1, y)$ must also be on the mirror giving you a second ...

2

One way to see that the best escape path must be straight is to note that once the stampede of buffaloes overtakes you, you can never overtake the stampede of buffaloes. Therefore you can get to a point on the perimeter of the square iff you can reach that point before the buffaloes. Now let's assume our square $ABCD$ is given by $[0,10]\times[0,10]$ and ...

2

HINT: If $(h,k)$ is the reflection of the point $(1,0)$ their mid point $\left(\dfrac{1+h}2,\dfrac{k+0}2\right)$ will lie on $y=mx$ and the line joining $(h,k);(1,0)$ will be perpendicular to $y=mx$ $$\implies m\cdot\dfrac{k-0}{h-1}=-1$$

1

Since according to your comment the rectangles are described by half their width and height, I assume that they are centered at the origin and aligned with the axes. So the smaller rectangle is $-w_1\le x\le w_1, -h_1\le y\le h_1$ and the larger likewise for $w_2,h_2$. Assume for the moment that $y=0$. Then the smaller rectangle is characterized by $\lvert ... 2 Observe that each vertex is shared by$6$faces, except the six vertices of the original octahedron, which are always shared by only$4$faces in each generation. Each face has$3$vertices. The number of faces in generation$N$is$8 \cdot 4^{N}$. Putting this together, we see that the number of vertices in that generation is $$\frac{3 \cdot 8 \cdot ... 3 The key is the Euler characteristic formula$$V-E+F = 2.$$You've already calculated that F=8\cdot 4^N. You also know that all of the facets are triangles; that means that each triangle has three half-edges or 3F=2E. That means$$2V - F = 4$$or$$V = 2 + 4^{N+1}.$$0 Assume without loss of generality that none of the segments are exactly horizontal. (If not, tilt your head slightly.) For a non-horizontal line segment, a line intersects the segment iff it passes below the segment's upper point and above the segment's lower point. The "above/below" relationship is preserved by duality in the following way: if a point is ... 3 Expand that expression and you'll see that it equals$$\lambda^2\,|A+B|^2 + (1-\lambda^2)^2\,|B|^2\,,$$which is clearly a non-negative quantity. 0 Consider the collection of vectors \{\vec x:\vec x\perp (3\vec i+\vec j+\vec k)\}. The endpoints of these form a plane through the origin. If you shift this plane upwards 7 units, you get the plane in question. 3 One way to think about this is to realize that a plane is made up of infinitely many parallel lines side by side. It's easy to think about lines in 2D space, and 3D space is just infinitely many 2D-planes laid vertically, infinitesimally next to each other. So when z=0, we would have the line 3x+y=7, which is just a line with y-intercept at (0,7) and ... 1 Purely by similar triangles, \frac{c}{d} =\frac{a}{\sqrt{a^2+b^2}} . No trig functions needed. 3 For a start, notice that$$c = d\sin\arctan\left(\frac b a \right)$$So$$d = c \sec\arctan\left(\frac b a \right) = c{\sqrt{1 + \left(\frac b a \right)^2}}$$Alternatively, if you compare angles, you can see that two of the triangles are similar, so$$\dfrac{d}{c} = \frac{\sqrt{a^2 + b^2}}{a} $$0 It will be easier to understand, perhaps, if we consider two vectors \vec{a} and \vec{b} restricted to the x-y plane, so that \vec{a} = a_i \hat{i} + a_j \hat{j}, and \vec{b} = b_i \hat{i} + b_j \hat{j}. Then \vec{a} \times \vec{b} = (a_i b_j - a_j b_i) \hat{k}, and \|\vec{a} \times \vec{b}\| = a_i b_j - a_j b_i. A geometrical ... 0 In general the plane: \color{blue}{ax+by+cz+d=0} has a normal vector having its direction ratios a, b, & c. The equation of the line having direction ratios a, b & c & passing through the point (x_{1}, y_{1}, z_{1}) is given by the general equation$$\color{red}{\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}=t} $$... 4 For simplicity of description, let us choose the coordinate system so that A is the origin, AB is in the direction of x-axis and C is in the upper half-plane. Let a, b, c be the length of sides BC, CA, AB. \alpha, \beta, \gamma be the angle of \triangle ABC at vertices A, B, C. The restriction on angles$$ \begin{cases} \angle YXZ ... 3 We may consider the following configuration: Given$XYZ$, we take$X'Y'Z'$as the anticomplementary triangle of$XYZ$. Then we take$\Gamma_A,\Gamma_B,\Gamma_C$as the circumcircles of$X'YZ,XY'Z,XYZ'$and$A\in\Gamma_A$. We take$B$as$AZ\cap\Gamma_B$and$C$as$BX\cap\Gamma_C$.$A,X,C$will be on the same line, and$\Gamma_A,\Gamma_B,\Gamma_C$will ... 1 This question is not that difficult to solve. (1) A = … = (1, 4), which will be used as the center of a circle. (2) Let$\theta$be the angle between the 2 given lines. Then,$\theta$or$\tan \theta$can be obtained by "the angle between 2 lines formula". (3) From (2), the value of$\sin \theta$is found. (4) Putting the 5 in the area formula "$Area = ...

0

The property of being perpendicular (perpendicularity) is the relationship between two lines which meet at a right angle (90 degrees). so try if $$x-y + z = 4;$$ then $$<x,y,z>$$ = $$<2,1,5> + t <1,-1,1>$$ which equals to $$<2+t , 1-t, 5+t>$$

21

1

I'll denote $[a\times b]$ or $[a,b]$ the cross product; $(a,b)$ or $a\cdot b$ the dot product and $(a,b,c)=[a\times b]\cdot c$. We're asked to find $$(CM,CB,BF)=[CM\times CB]\cdot BF=[CB+BM\times CB]\cdot BF= ([CB\times CB]+[BM\times CB],BF)=(0+[BM\times CB],BF)$$ Now, I'd suppose "The point $M$ divides the segment $AB$ on ratio 2" is $\frac{AM}{MB}=2$, as ...

0

I assume that the four given points are as given as four trees in a landscape forming a nondegenerate rectangle. If your question could not be decided this would mean that we have four different circles intersecting in at least two different points $P$, $Q$. But this would imply that the four circles all have their center on the median $m$ of $P$ and $Q$, ...

0

I think that the formula I gave is correct. I did some numerical test for different dimensions and I was able to recover the correct number of points on the ($N-1$)surface.

1

For any point $X$ inside the rectangle, except the rectangle's center, there exists another rectangle, which has the same vertices-to-X distances, but does not contain $X$. You can construct it by mirroring one of the rectangle's edges by an axis parallel to the edge and passing through $X$. The opposite is also true: you can make a reverse construction for ...

4

It's not possible in general: A . . . . . D . . A' . . . . . . B . . . . . C . . B' Let $X$ be the centre of $ABA'B'$. Then $XA=XA'$ and $XB=XB'$, and of course $XC=XC$ and $XD=XD$. But for suitable choice of $C$ and $D$, $X$ is inside $ABCD$ but not $A'B'CD$.

8

It seems it is impossible to know whether X is inside for sure. For example, if we had a rectangle ABCD and a point X inside of it, WLOG so that the vertical distance from X to CD is less than the vertical distance from X to AB, creating a new rectangle ABEF with EF parallel to AB and CD, but EF underneath X with the same vertical distance creates a new ...

2

I am giving below a rather simple solution using vectors. Please make your own figure as I could not put in the same. Also treat small letters $a$, $b$, and $c$ as vectors. Write to me if you need more explanation. Let $ABC$ a be triangle, let the perpendicular from $A$ to $BC$ meet $BC$ at $D$, and let the perpendicular from $B$ to $AC$ meet $AC$ at $E$. ...

2

Even though programmers know that the term can be applied when placing data in one, two, or multiple dimensions, I believe that "array", if used in elementary school, gets used specifically to refer to a two-dimensional arrangement of things. I certainly remember a similar experience growing up: after being exposed to the idea of arrays being ...

0

Programmers in particular will call them arrays as that has been the term in all the languages I have worked with. This does not bring attention to the fact that all your examples have exactly two dimensions and the dimensions are the same. Arrays can have any number of dimensions and each dimension may be a different size. As figures, I think squares is ...

-1

The rectangular arrangement of the square boxes is meant by the term "array". The German word for this would be back translated to "field", like in grainfield. The geometric relation might be visible from the Japanese character for field: 田 If you look in an English dictionary on the origin of words (etymology), you get: From Middle English arrayen, ...

-1

You are correct. The photos that you posted are examples of multidimensional arrays. It makes sense as well since the definition of an array is an arrangement of objects that can be constructed in rows and columns.

3

Yes. The interior angle is always double the inscribed angle (in your case $D$).

2

One side of this, there is a standard picture. In $\mathbb R^n,$ take the $n$ points $$(1,0,0,\ldots,0),$$ $$(0,1,0,\ldots,0),$$ $$\cdots$$ $$(0,0,0,\ldots,1).$$ These are all at pairwise distance $\sqrt 2$ apart. At the same time, they lie in the $(n-1)$-dimensional plane $$x_1 + x_2 + \cdots + x_n = 1.$$ If you wish to work at it, you can ...

1

For the point $(x+s\cos(\theta),x^2+s\sin(\theta))$ to be on the curve $y=x^2$, we need $$(x+s\cos(\theta))^2=x^2+s\sin(\theta)\tag{1}$$ Expanding the left side of $(1)$ and cancelling, we get $$s\cos(\theta)=\tan(\theta)-2x\tag{2}$$ Now $(2)$ must also remain true for $\theta+\frac\pi2$ so that the other corner adjacent to $(x,x^2)$ is on the parabola. ...

0

Loosely speaking: The error comes from problems at the boundary. If you enlarge $S$ to contain all partial hexagons, then the $\frac\sigma n$ becomes exact. Since the partial hexagons live in a strip of thickness $\sim \frac1{\sqrt n}$, we end up with an error of $O(\frac1n\cdot \frac1{\sqrt n})$

0

The angle you traveled is $\theta=2\pi\cdot100/40000=0.0157$ radians. Then I walked $H=R(1-\cos(\theta))=40000/2\pi\cdot(1-\cos(0.0157))=0.7854$ km. (Check: with $10000$ km, $\theta=\pi/2$ radians, $H=40000/2\pi\cdot(1-0)=6366.2$ km.)

1

you should edit your question explaining what you did in comment. anyway, take a look at Cartesian to Radial coordinate transformations. https://en.wikipedia.org/wiki/List_of_common_coordinate_transformations

0

A direction vector of the straight line given is $(1,1,3)$. We are on the surface if $(x,y,z)+(u,u,3u)=(t,\frac{t^2}{2},0)$ for some $t$ and $u$. Intuitively: sitting on the surface (at $(x,y,z)$) and stretching a vector (parallel to, say $(1,1,3)$) towards the $xy$ plane, the vector has to reach it at a point for which the directrix equation holds: We ...

4

Any two Riemannian manifolds with constant sectional curvature $C$ are locally homogeneous (in normal coordinates, one has an explicit description of the metric and by composing two normal coordinate systems around two different points one obtains a local isometry). However, such spaces need not be homogeneous. For example, consider a closed oriented ...

1

More generally, any reducible variety is either disconnected or singular (or both!). This is because if there are two components that intersect nontrivially, a point in the intersection would not have its local ring be a domain, much less a regular local ring. Now you have a reducible cubic surface which is necessarily not disconnected because two ...

1

In this case: no $z\implies$ is an (hyperbolic) cylinder $\implies$ obviously a ruled surface.

1

For the first question, the answer is yes. We can choose a point $P$ on the nonsingular plane curve $C$ of degree 4. Project $C$ from $P$ onto a hyperplane (i.e. a copy of $\mathbb{P}^1$) in $\mathbb{P}^2$. Noting that any line passing through $P$ cuts $C$ at other three points, this projection is a rational map of degree 3.

1

Let $r_1,r_2$ be respectively the radii of the first and second circles, then: $r_1+r_2 = 11, \pi(r_1^2+r_2^2) = 230$. Can you move forward from here?

2

I think you are looking for the volume of an $n$-dimensional simplex. If you have $n+1$ points in $n$ dimensions (where the points are given as vectors $v_0,v_1,\dots,v_{n}$), then the formula for the volume is given by $$\left|\frac{1}{n!}\det\left[\begin{matrix}v_1-v_0\\v_2-v_0\\\vdots\\v_{n}-v_0\end{matrix}\right]\right|$$

0

The radius $R$ of the spherical surface passing through all 20 identical vertices of a dodecahedron with edges length $a$ is given by generalized formula (derived in HCR's Formula for platonic solids)$$\bbox[4pt, border: 1px solid blue;]{R=\frac{\sqrt{3}(\sqrt{5}+1)a}{4}}$$ $$\implies a=\frac{4R}{\sqrt{3}(\sqrt{5}+1)}$$\implies ...

Top 50 recent answers are included