# Tag Info

## New answers tagged geometry

0

Restricted to any plane perpendicular to the line of intersection of the reflection planes, the reflections are a pair of orientation reversing isometries fixing the same point. Their composition is thus an orientation preserving isometry fixing one point: a rotation. To see what the rotation angle must be, track the effect of each rotation on angles ...

0

Oblique cylinders have the same height as regular cylinders in terms of $r$ and $h$! To convince yourself, try this, and you'll see that an oblique cylinder is just a warped right cylinder. So if the slant angle is $60$ degrees, the height is $3\sqrt{3}$ by the $30-60-90$ triangle properties. So then the volume is $\pi r^2 h=\pi (9)^2(3 ... 1 Write$c := \cos\theta$,$s := \sin\theta$,$\mathbf{w} := \mathbf{u}\times\mathbf{v} = s\mathbf{n}$, and$T := T_\mathbf{u}\left(T_\mathbf{v}\right), so that we have ... \begin{align} T(\mathbf{x}) &=\mathbf{x}-2(\mathbf{x}\cdot\mathbf{u})\mathbf{u}-2(\mathbf{x}\cdot\mathbf{v})\mathbf{v} + 4c(\mathbf{x}\cdot\mathbf{v})\mathbf{u}\\ R(\mathbf{x}) ... 0 A fuzzy handwaving argument (which often works) ... The problem is "symmetric" in the three points. There is nothing special about any one of them; they all have equivalent roles and status. Therefore the solution must also be symmetric, in the same sense. So, the triangle has to be equilateral. 2 Hint: Fix any two points and draw the chord AB. Now the area is maximised if the third point C is farthest from AB, i.e. on the perpendicular bisector and when the triangle is isosceles. As we can choose the first two points arbitrarily, the triangle has to be in fact equilateral. 1 I actually found a solution to this problem. If I'll be able, I provide some images, thank you everyone for help! 1 Let the side length be l. All sides are equal, so l \times l \times l=l^3=729. Solve for l. 1 Let \{v_{i}\}_{i=1}^k be a set of k vectors in \mathbf{R}^n. By "sum of all pairs of inner-products", presumably you mean something like \sum_{i<j} \langle v_{i}, v_{j}\rangle, $$and by "sum of Euclidean distances between all pairs" you mean$$ \sum_{i<j} \|v_{i} - v_{j}\|. $$Consider what happens with two vectors. Since you're asking about ... 1 The distance from \bar{D} to \bar{BC} is the perpendicular line from \bar{D} to \bar{BC}. This is:$$DC\sin (m\angle{C})=8\sin (m\angle{C})=\frac{32\sqrt{5}}{9}$$We draw an auxilliary line from B perpendicular to \bar{DC}. Let M be the point where it meets \bar{BC}. The \triangle BCM is a right triangle with side lengths 1, 4\sqrt{5}, ... 0 You can also build it on the Axiom that there is one and only one line between any two distinct points in space. Suppose two distinct lines l_1 and l_2 intersect at more than one point. Pick any two of these points. Both l_1 and l_2 are lines which join these two points. But thiis contradicts the fact the two lines are disjoint proving there can ... 1 As long as you are not obliged to use your formulae for T_{\bf u}({\bf x}) and so on, there is a very easy approach to this. Any reflection in {\Bbb R}^3 is represented by an orthogonal matrix of determinant -1. Any rotation in {\Bbb R}^3 is represented by an orthogonal matrix of determinant +1. Multiply two of the former and you get one of the ... 2 Yup: it states that$$ a=\sqrt{b^2+c^2-2bc\cos\alpha} $$in stantard trigonometric notation (where a,b,c are the sides of the triangle, and \alpha is the angle which is opposite to a). A simple proof can be given with vector calculus: being \vec{a}=\vec{b}-\vec{c}, squaring this relation you obtain:$$ \vec{a}\cdot\vec{a}\equiv a^2 = b^2 + c^2 - 2 ... 0 Since you specified the other side is7$, then: $$a^2+7^2=25^2$$ $$a^2+49=625$$ $$a^2=576$$ $$\boxed{a=24}$$ The answer to a) is$a=24$. If you have to do it with trigonometry, let$\theta$be the top angle (between the$7$and the$25$). Then: $$\cos\theta=\dfrac{7}{25}$$ $$\theta=\cos^{-1}\left(\dfrac{7}{25}\right)$$ Now you know the angle measure of ... 1 @ user133921 : Even is your result is apparently correct, your calculus is wrong : there is a mismatch with$a$and$b$. You find$a=66$and$b=114$. Hence$a-b=-48$which doesn't agree with the given equation$a-b=48$. I suppose that you will find by yourself where is the mistake. 0 Let the measures of the angles be$a$and$b$. Also, let$a$be the larger angle ($a > b$) $$a+b=180^\circ$$ $$a-b=48^\circ$$ Subtract the first equation from the second to get rid of$a$. This is called elimination. $$2b=132^\circ$$ $$b=\dfrac{132^\circ}{2}$$ $$b=66^\circ$$ Plug the value of$b$into either equation. $$a+66^\circ=180^\circ$$ ... 0 Using the forms we have separated the notion, the function, from its arguments. For example, the volume form calculates the signed volume, a number, of a given n-vector. It is useful in curved spaces, where the forms take "infinitesimal" vectors (at a given point), hence the name differential form (field). 0 This problem is discussed in Section 11 of Geodesics on an ellipsoid of revolution (it's the "trilateration problem"). 0 Here's another resource Geodesics on the Torus and other Surfaces of Revolution .... 0 You don't have to do integrals! Divide atmospheric pressure A = 101.3 kPa by g = 9.8 m/s2 to give the mass per unit area (kg/m2). Multiply this by the area of the earth and you're done. (Assumptions: g is a constant over the height of the atmosphere; g independent of latitude; neglect the mass of the air displaced by the volume of the land about sea ... 0 It just a transverse Mercator projection. The library GeographicLib (written by me) has a class OSGB which converts eastings + northings to latitude + longitude (and also deals with OS grid references). If you want to know how to calculate the transverse Mercator projection accurately, see Transverse Mercator with an accuracy of a few nanometers (a ... 1 There is an excellent account on this Wikipedia page. The first two-thirds of the article cover ellipsoids of revolution (which is what you have). 0 There is an excellent description of geodesics on a torus in this document. 3 That animation is custom-made in Flash, which is a perfect and full-blown tool to do 2D animations(and even games) like that but it has a quite heavy cost, even for the student license($\$19.99$ per month or $\$199$per year). GeoGebra is a free tool for drawing geometry, and it has capabilities for doing interaction, animations, and it has a beta for 3D ... -1 First of all you need to know what is the connection you want to compute the geodesics of. If you want a metric connection you must know the metric at play, which in most cases will be the one induced by embedding in$R^3$. Once you know that you can solve the geodesics equation with respect to appropriate coordinates and find an analytic expression for ... 0 Hint: for Pick's Theorem ($AREA = I+\frac{B}{2}$-1) if there are more integer points contained in the polygon ( I ), then there are more chance to find some rectangles...so you have to place the polygon in order to have the least number of lattice points that are also in the polygon's perimeter ( B )... 0 The locus of points for which the angle between the monuments is fixed is a circle. If you have three monuments, you have three pairs, and thus three circles, which all intersect at your location. 0 You are basically correct, although I think if you have two monuments then you have narrowed it down to two locations (one on either side of the two monuments). Knowing a third helps as long as these points are not colinear. 0 I guees you know how to express lenght of medians in function of lenght of sides. Have you tried working with trigonometry expresions? Replacing$a=2R\sin \alpha$and$S=\frac {ab\sin \gamma} 2$? 0 Parent rectangle horizontal axis of symmetry lies at$26/2=13$units from O and so should the horizontal axis of symmetry of the child rectangle. Parent rectangle vertical axis of symmetry lies at$305/2=152.5$units from O and so should the vertical axis of symmetry of the child rectangle. Drawing these two lines intersecting (at say, O') on the parent ... 0 If the big rectangle is$a \times b$(in your diagram$a=26, b=305$), and the smaller rectangle is$c \times d$(here$(c,d)=(18,68)$), then your$x,y$satisfy$2x+c=a,2y+d=b$so that$(x,y)=((a-c)/2,(b-d)/2).$A check:$(x,y)=(4,118.5)$so the check would be$4+18+4=26$and$118.5+68+118.5=305.$0 I think the reason people identify these$k$-forms with geometric objects is because of integration. People see$ dx^1, dx^2, \ldots$in integrations and think that they represent the manifold being integrated over, that basis forms are what you need in order to write integrals. But this is not true at all! An integral of a$k$-form-field involves ... 3 You can cover the plane in rectangles if you want, in a checkerboard pattern - alternating copies of the original and reflections of the original. You can use this to find the shortest path by locating the two points in each - reflecting in one side at a time. Sometimes this is used in theoretical snooker, billiards or pool with tables which have perfect ... 2 I will prove the following: Let$ABCD$a rectangle having$|AB|=a$and$|BC|=c$and let$K$,$L$,$M$,$N$be points on$AB$,$BC$,$CD$and$DA$, respectively. Then the perimeter of$KLMN$is at least$2 \sqrt{a^2+b^2}$. This is achievable by letting$K$,$L$,$M$and$N$the midpoints of the sides of$ABCD$. Write$k = |KB|$,$l = |LC|$,$m = |MD|$,$n = ...

3

Unflod your rectangle to make a grid and place K, L, M and N on it. You goal is to shorten the distance $KL+LM+MN+NK'$. You need to align $KLMNK'$ to reach the minimal distance for $KK'$. That's $2.AC$ Here is a picture of a non-optimal situation. If it were optimal, then $KLM'N'K'$ would be a straight line. Here is a picure of an optimal situation:

1

Here's a method to compute the distance (in km along the surface of the earth) between two points whose longitude and latitude (in degrees) are given. Assuming the earth is a perfect sphere of radius $R$ km, introduce the Cartesian coordinate system with origin at the center of the earth, $x$-axis through the intersection of the equator and the prime ...

1

Brainstorming answer to a brainstorming question… I'm not happy with this answer myself, and hope that me posting this answer doesn't prevent someone likely to give a better answer from reading your question in the first place. You could start by identifying two points with maximal distance from one another, which you can obtain in linear time from the ...

0

Don't know if a proof with induction meets your requested answer... But by induction this (seems) to be pretty easy to show: Proof: Let $P = \{P_1,\ldots,P_n\}$ be a set of $n$ distinct points in $\mathbb R^2$ with the precondition, that $\forall a,b,c \in \{1,\ldots,n\} \exists d \in \{1,\ldots,n\}$ such that $P_a,P_b,P_c,P_d$ are con-cyclic. (Induction ...

2

$2*BC = BC + BC$ $= BC + AC$ (tangents from a common point to a circle are equal common-point C ) $= BF + FC + CD + AD$ $= EF + FC + CD + DE$ (tangents from a common point to a circle are equal ) $= CD + FC + DF$ $=$ Perimeter of the triangle

1

Hint: You said using geometry, not calculus, so integration should not be required. Can you figure out what the shape is? You might try a different coordinate system.

0

Cartesian coordinates are isothermal for your metric, so writing $\lambda = 1/f(r)$ you have the standard formula $K = -(\Delta \log \lambda)/\lambda^{2}$. (The computation is indeed a bit messy, but the formula comes out as stated.)

2

http://www.math.uh.edu/~minru/4350-11/geodesic.pdf See pages 3 and 4 of this PDF!

1

Here's one way to look at this: Suppose $\nabla$ is the unique covariant derivative operator associated with the metric tensor $g$ on the surface $S$, i.e. it is the unique symmetric connection on $S$ with $\nabla g = 0$. It is well know that the geodesics $\gamma(t)$ of $g$ satisfy $\nabla_{\mathbf w}\mathbf w = 0, \tag{1}$ where $\mathbf w$ is the ...

1

You are looking to scale down a triangle by dividing each side by some value $x$ such that $$\frac Ax+\frac Bx=\frac{A+B}x=1$$. It quickly follows that $x=A+B$.

0

Given your triangle $ABC$, you are looking for a triangle $A' B' C'$ such that $A' + B' = 1$ and you want to preserve similarity such that $\frac{A + B}{C} = \frac{A'+B'}{C'}$ which is equal to $\frac{1}{C'}$. So you've got three equations and three unknowns :)

6

Riffing on @Shuchang's answer ... Starting with unit circle $\bigcirc O$, one easily constructs $A$, $B$, $C$, $D$ with $|\overline{AB}| = \sqrt{2}$ and $|\overline{CD}| = \sqrt{3}$. Quadrisecting $\overline{AB}$ and $\overline{CD}$ one draws $\bigcirc A$ and $\bigcirc{B}$ to provide chords of length $\sqrt{2}/4$ and $\sqrt{3}/4$. Chains of congruent ...

1

A more analyitc solution can be this one: the equations of the angle bisectors of the angle created by 2 lines whose implicit equations are $$a_1x+b_1y=c_1 \qquad(1)$$ $$a_2x+b_2y=c_2 \qquad(2)$$ can be found as $$\left( a_{1}\sqrt{a_{2}^{2}+b_{2}^{2}}-a_{2}\sqrt{a_{1}^{2}+b_{1}^{2}}% \right) x+\left( b_{1}\sqrt{a_{2}^{2}+b_{2}^{2}}-b_{2}\sqrt{% ... 1 Think in vectors. Start with two difference vectors$$ \overrightarrow{ST}=\overrightarrow T-\overrightarrow S \qquad \overrightarrow{SU}=\overrightarrow U-\overrightarrow S $$Scale them so that the length becomes one:$$\frac{\overrightarrow{ST}}{\lVert\overrightarrow{ST}\rVert} \qquad \frac{\overrightarrow{SU}}{\lVert\overrightarrow{SU}\rVert}$$Add ... 11 This is what I'm trying to show on a diagram indicating explicitly all quantities and the approximation is quite rough. Here is a circle with center O. Quantities OA=OB=OC=OD=AB=1 and OC\perp OA, OD\parallel OA. Line segments AC and BD have an intersection E. We can easily deduce the following quantity.$$AC^2=OA^2+OC^2=2\qquad ...

0

if slopes of lines with any two point will be same , then they are co-linear i.e. $$\frac{y_2-y_1}{x_2-x_1}=\frac{y_3-y_1}{x_3-x_1}$$

0

Equation of a straight line is a linear function ($y=ax+b$) now if the system $$y_1=ax_1+b\\y_2=ax_2+b\\y_3=ax_3+b$$ has solutions it works,otherwise not

Top 50 recent answers are included