# Tag Info

0

Well, by the law of sines, you have $$\frac{a}{\sin(60^\circ)}=\frac{4}{\sin(x^\circ)}$$ where $x^\circ=\angle QRP$, and we have $0<x<120$ (and $a>0$). To find how many triangles there are, all we need to do is to count the number of solutions for $x$ in this equation. First we rewrite the equation be evaluating $\sin(60^\circ)=\frac{\sqrt{3}}{2}$ ...

1

I asked the same question to myself a while ago, and now I'm glad to share my conclusion. To build a triangle given $O,H,I$ is substantially equivalent to solving a general cubic, hence it cannot be done with straightedge and compass only. To prove it, notice that we know $$(R,r,a^2+b^2+c^2)$$ and we have to find $(a,b,c)$ or, equivalently, the exradii ...

1

Certainly, an abuse of notation happened here. Let's see if illustration will help (honestly, I've tried to do my best to draw it as pleasantly as possible): So, it's confusing that curve has the same name as it's codomain. What was really meant is that you can parameterize a patch of hypersurface with neighbourhood of euclidean space of lesser dimension ...

0

A surface is a space whose points have each an open nghb homeomorphic to an open disc in $\mathbb R^2$. This is equivalent to have an open nghb homeomorphic to an open set in $\mathbb R^2$: the latter always contains an open disc! But the problem is that one cannot say $U=S\cap B$ is homeomorphic to an open disc, only that it contains some open set $W$ of ...

0

choose the coordinates so that the equation of the parabola is $y = x^2.$ let the triangle be $OAB$ with $O =(0,0), A = (a, a^2), B= (b, b^2)$ and $\angle AOB = 120^\circ$ by cosine rule, $$AB^2 = OA^2 + OB^2 + OA*OB$$ so that $$(a-b)^2 + (a^2 - b^2)^2 = a^2 + a^4 + b^2 + b^4 + \sqrt{(a^2 + a^4)(b^2+b^4)}$$ cleaning it up a bit we end up with $$3a^2b^2 + ... 0 The set Y=\{f=0\}\subset X is analytic, hence it has countable collection of irreducible analytic components. Furthermore, those components form a locally finite familly (this is basic Complex Analytic Sets theory, e.g. Gunning-Rossi, Narasimhan...). Now any analytic set H\subset Y of codimension 1 in X would be an irreducible component of Y. Hence ... 0 Let \mathbf{v} be a unit vector along the common axis of the cones, let \mathbf{a}, \mathbf{b} be the vectors joining the vertex to the points on the yellow and blue cones' bases. Since we know the radii and the height of these cones, we can compute \theta_1, \theta_2 the angles between \mathbf{v}, \mathbf{a} and \mathbf{v}, \mathbf{b} ... 1 If by 120 you mean 120^\circ  then it must be on the vertex of parabola. Proof is easy. Just take any point A on parabola, which is not vertex (let vertex point be point V) of parabola, then draw line from this point to vertex, then draw line from this point, such that it will form angle with 120^\circ . You will get point B, which is intersection of ... 0 So, we have f(x) = ax^2 + bx + c, where a is distinct from 0. 'A' is related with growth. If 'a' is negative, function decrease, if 'a' is positive, function increase. 'C' is of course y-argument of point (0, f(x)) (let's imagine this as a place where function cross OY axis). 3 For any varieties over any field, you can associate the group of Weil divisor, that is the free abelian group generated by the codimension 1 irreductible subvarieties. It is denoted Z^1(X). Now, if f is a rational function and V a subvariety of codimension 1, there are many ways to define the order of f=a/b along V, some of them work only under ... 0 Let the angle rotated be z about the fixed focus and lets take the standard equation of ellipse Use the parametric form of line passing through the fixed focus and with slope \tan z to find the point through which the nearer directrix passes like, (x-ae)/\cos z = (y-0)/ \sin z = -r ,where r= a/e -ae find x and y and hence the equation of ... 1 Suppose f:=\begin{bmatrix}f_1\\f_2\end{bmatrix} is one of the foci. And \theta is the angle you want to rotate. For every point x:=\begin{bmatrix}x_1\\x_2\end{bmatrix} of the ellipse do this. x-f=\begin{bmatrix}x_1-f_1\\x_2-f_2\end{bmatrix} This brings the focus f to the origin. ... 0 I don't think it has a special name. the line is not so special. (every point has one, two of these lines from different points but the same axis can intersect, while if they are perpendiculatr to the same line they don't) Maybe best is to describe it as: "the line trough P that is perpendicular to the perpendicular from P to l" or "the line ... 0 If "perimeter" is the lenght of..., is impossible, as already answered. But if the form is know is possible calculate the area using Green's theorems. The Planimeter is an application of this fact. 0 Use 60.8926 to obtain \theta, the semi-central angle (approximately equal to 79.21^0). Use power of a point to find R (i.e. 18(2R - 18) = 21.5^2, such that R = 21.8427778) Consider \triangle OCG to find d. [R - 18 = (R - d) \cos \theta] Use \triangle KFO to get the required. [\angle FKO = arctan \frac {(18 - d)}{21.5}] 1 It is well-known fact that these three points are collinear. They lie on so-called Simson line of point P with respect to triangle ABC. I think you can use a Menelaus' theorem here. Assume that P_1 \in BC, P_2 \in CA and P_3 \in AB. Note that \frac{BP_1}{P_1C} = \frac{\cot \angle PBC}{\cot \angle PCB} and so on. After multiplying this equality ... 2 The Simson theorem can be proved by angle chasing only. \hspace2in Since BP_1 P P_2 is a cyclic quadrilateral,$$\widehat{BP_2 P_1}=\widehat{BPP_1}=\widehat{ABP}-\frac{\pi}{2},$$and since CPP_2P_3 is a cyclic quadrilateral,$$\widehat{CP_2 P_3}=\widehat{CPP_3}=\frac{\pi}{2}-\widehat{PCA},so \widehat{BP_2 P_1}=\widehat{CP_2 P_3} proves that ... 3 Contrary to all other answers, I say yes you can find the area a of a known shape (clover leaf) from the length of it perimeter p. Taking a similar model of the clover leaf, measure its area A and perimeter P, using a curvimeter and a planimeter. You can also do that from a digital image (photoscan), but I don't know of ready-made tools for that. ... 1 You can think of your coordinates as a parameterization {\bf r} of the appropriate subset of the xy-plane in the coordinate. Then, in the coordinate basis (x, y), \begin{align} \partial_{\kappa} &= {\bf r}_{\kappa} = (\mu, -\kappa) \\ \partial_{\mu} &= {\bf r}_{\mu} = (\kappa, \mu) . \end{align} 1 No, this is not possible, because you can easily increase the perimeter of any polygon (while retaining its area) by modifying its edges like this: (Source: http://en.wikipedia.org/wiki/File:Quadratic_Koch_curve_type2_iterations.png) You can even use this technique to create fractals with an inifinite perimeter, such as the "Minkowski Sausage". PS: As ... 1 A hint: The point O can be written in two ways: Since O is the intersection of the two lines L\vee N and K\vee M there are uniquely determined s, t\in[0,1] such thatO=(1-s){A+B\over2}+s{C+D\over2}=(1-t){B+C\over2}+t{D+A\over2}\ .$$Here s and t can immediately be guessed. 2 Yes, the diagonals meet at a common point. Certainly, the diagonals are parallel to, and two-thirds the length of, corresponding sides of the triangle. Consider a sub-triangle (say, \triangle AF_2 E_1) created by one of the diagonals and an appropriate vertex. The other diagonals (F_1D_2 and E_2 D_1) are parallel to sides (AF_2 and AE_1) of that ... 1 they do meet at the center of the triangle. to see this let we a, b, c represent the points. call the points A_1, A_2 on BC such that BA_1 = A_1A_2 = A_2C the point a_1 = 2/3 b + 1/3 c, a_2=1/3b + 2/3 c similarly define points b_1=1/3 c+2/3a, b_2 = 2/3 c+ 1/3 a. the point where all diameters meet is mid point of A_1B_1 given by (a+b+c)/3. 3 First of all, welcome to Math Stackexchange! There is no such thing as a single vector representation over all of these four points. Mathematically, a vector in the Euclidean \mathbb{R}^n is a tuple of n numbers — for example$$\begin{pmatrix}a\\b\end{pmatrix},\quad a,b\in\mathbb{R} $$If you consider the 2-dimensional case, just as in your question. ... 1 In general, no, though the Isoperimetric Inequality gives and upper limit on area: For a bounded region in the plane with perimeter L, the area A must satisfy$$A \leq \frac{L^2}{4 \pi},$$and equality holds iff the region is a circle. By constrast, if the region is a square, we have A = \left(\frac{L}{4}\right)^2 = \frac{L^2}{16}, which (since \pi ... 1 The second option would be upper-bounded by the first. See the triangle inequality. 2 Perimeter is basically arc length. You could just as easily have a line segment of length 96 as a square of perimeter 96. Then the side length is 24, and so the area is 24^2. On the other hand, if the circumference of a circle is 96, then 96=\pi d, so that means that r=96/ 2\pi. This means that the area is (96/2)^2, which is different from 24^2. 1 This is impossible. We can prove this by constructing two shapes with the same perimeter but different areas. Consider, for instance, the unit equilateral triangle, with perimeter 3 and area \sqrt{3}/4, and the square with side lengths 3/4 and area 9/16. Since these two shapes have the same perimeter, but different areas, one cannot uniquely ... 0 Maybe for that problem the short path is seeing the triangle as the half of an equilateral triangle. Therefore 20 is the height and you know the relationship between the size and the height:$$h=s\frac{\sqrt{3}}{2} \implies s=\frac{2h}{\sqrt{3}}=\frac{40}{\sqrt{3}}$$0 I'd put equation 2 into the form y = mx + b. 10x + cy = 8 cy = -10x + 8 y = (-10/c)x + 8/c For the slope of equation 2 to be perpendicular to equation 1, the following most hold: -1/(-10/c) = 5 A slope m_2 is perpendicular to a slope m_1 whenever m_2 = -1/m_1. So what we did above is we found the slope that is perpendicular to the slope ... 0 hint: two lines are perpendicular if the product of their slopes is negative one. can you find the slopes of y = 5x -6 and 10x + cy = 8? 0$$a\perp b$$if the slopes are the negative reciprocals of each other.$$10x+cy=8\implies cy=8-10x \implies y= \frac{-10x}{c}+\frac{8}{c}What value of c would make the x term have a coefficient of -\frac{1}{m_1}= -\frac{1}{5}? 1 If you're trying to find mathematical sense in the physical world, then let me ask you this: How do you measure length? Do you count the number of molecules? atoms? quarks? Tell me what you basic unit is, and I will ask you to represent \frac12 with it. By the way, these numbers are called irrational because they don't make (rational) sense. In ... 1 WOLOG, choose the coordinate system such that the sphere of radius a is centered at origin O and the axes of the two holes are aligned along the x and y axis. Letc = \sqrt{a^2 - b^2},\quad d = \begin{cases}\sqrt{b^2 - c^2},& b > \frac{a}{\sqrt{2}}\\0,& \text{otherwise}\end{cases} \quad\text{ and }\quad e = \min(b,c) $$When one ... 2 The important fact here is that you can't have EXACLTY a 1 meter slab because there is (even if very small) an "uncertainty" with the length of the slab. In fact we could measure the length of the slab infinitely many times but it won't never be 1 exactly (for more details see this numberphile video). So if we can't have a length 1 we can't have ... 0 Yes, you can treat the points on a line as the set of all linear combinations of the two defining points, the way you did. In my nomenclature, I'd usually use different vectors to describe the line itself. In \mathbb P^2 I'd use the cross product to obtain a vector normal to all the points. In \mathbb P^3 I'd use Plücker coordinates. But there are ... 0 I have attached images with equations in word format 0 Some observations A is on the plane, (2,1,0) is on the plane, (0,4,-3) is the normal of the plane, hence the line is perpendicular to the plane. Then d(A,r) becomes the distance between A and (2,1,0), which is easy to calculate. And to find C which satisfies the requirement, it's enough to find a point C on r (whose coordinate can be ... 0 You can compute this as$$d=b\,E\bigl(\tan^{-1}(a/b\,\tan(\theta))\,\big|\,1-(a/b)^2\bigr)$$using the incomplete elliptic integral of the second kind E(\varphi\,|\,m). In Mathematica-Syntax (and suitable for Wolfram Alpha) this can be written as 2.23*EllipticE[ArcTan[3.05/2.23*Tan[50°]],1-(3.05/2.23)^2] I adapted this from this post which ... 0 We may assume that the sphere is given by$$S:\quad x^2+y^2+z^2=a^2, \qquad a>0\ ,$$and the cylinder by$$C:\quad (x-m)^2 +y^2=b^2,\qquad m\geq0, \quad b>0\ .$$The sphere intersects the (x,y)-plane in the circle K_s:\ x^2+y^2=a^2, and the cylinder in the circle K_c: \ (x-m)^2+y^2=b^2. The vertical projection of the curve \gamma:=S\cap C is a ... 4 There are at least four relevant articles of Beltrami. In the first one (Ann. Mat. Pura App.I (1865), no. 7, 185-204) he computes all two dimensional Riemannian metrics for which a coordinate system exists where all geodesics are represented as straight lines: he finds out that only surfaces of constant curvature have this property. I have little doubts that ... 1 You just have to add the rotation angle to the angle you calculate with 2i\pi / n BTW, your x & y are the opposite of the conventional, which will reflect your polygon around the y = x line, effectively rotating it. So you want$$\begin{align} x & = r \cos(\alpha + 2i\pi / n)\\ y & = r \sin(\alpha + 2i\pi / n)\\ \end{align} $$where ... 3 Since the question itself has already been answered, this answer will add some references for further study. As Joseph Zambrano points out in the comments to his answer, "local rigidity" fails for two dimensional hyperbolic structures. This is a hint of the deep, beautiful theory of moduli spaces of geometric structures. Bill Goldman's survey paper ... 0 here is how you will do. any point on the line connecting (x_1, y_1) and (x_2, y_2) can be written in the form$$x = tx_2 + (1-t)x_1, y = ty_2 + (1-t)y_1$$for the point (x,y) to be on the unit circle, t needs to satisfy$$ \left(tx_2 + (1-t)x_1 \right)^2+ \left( ty_2 + (1-t)y_1\right)^2 = 1$$follow these steps: (a) solve the quadratic equation ... 3 I shall assume that your$$\gamma'(t)=\bigl(x'(t),y'(t)\bigr)\qquad(a\leq t\leq b)$$is continuous. Furthermore it is no restriction of generality to assume$$P=(0,0),\qquad Q=(q,0),\quad q>0\ .$$Then$$\eqalign{\int_a^b\|\gamma'(t)\|\>dt&=\int_a^b\sqrt{x'^2(t)+y'^2(t)}\>dt\cr &\geq\>\int_a^b\bigl|x'(t)\bigr|\>dt\geq\int_a^b ...

2

In principle the integral is not too difficult using polar coordinates. I will do everything for $x_0=-x_1 ,y_0 =- y_1$ and $p=q=0$ to simplify notation a little bit. Then we don't have to shift and the whole area is given by $4$ times the integral over the first quadrant. We split the remaining integral in two regions. 1.) The triangle with vertices ...

2

Actually, you can't. Maybe some background information is of interest. What you found is the special case of a parametrization parametrized 'proportional to arclength', that is the length of a segment of the curve is a constant times the difference of the parameter values. (That means that you can show the minimizing curve is a straight line, but you cannot ...

1

This is basically Langley's Adventitious Angles problem, a problem that first appeared in a $1922$ Mathematical Gazette, and is well known for being extremely difficult despite its simplistic appearance. Here is a nice piece by piece solution with diagrams. For the sake of completeness, I will go ahead and add the Wikipedia solution which is attributed to ...

2

First, observe that $\angle DTE = \angle DTJ + \angle JTE = \left(\pi - \angle JAD\right) + \left(\pi - \angle EBJ\right) = 2\pi - \angle JAD - \angle EBJ = 2\pi - \left(\frac \pi 2 + \frac \alpha 2\right) - \left(\frac \pi 2 + \frac \beta 2\right) = \pi - \frac \alpha 2 - \frac \beta 2.$ On the other hand, $\angle EKD = \pi - \angle DKA - \angle BKE = \pi ... 0 Draw the circle passing through$B,C,D$and the circle passing through$A,C,D$(or the line for three collinear points). The argument of the cross-ratio$\arg(A,B,C,D)$is the angle between the two circles* where they meet at$C\$. To work this out, notice that the construction and the answer you get are invariant under Möbius maps, so you can make ...

Top 50 recent answers are included