New answers tagged

0

Alright, I will switch to Lehman's notation; here we have a quadratic form in three variables $x,y,z$ written as $$ g(x,y,z) = a x^2 + b y^2 + c z^2 + r yz+szx + t xy. $$ At the end, you would want to set $z=1$ and set the mess to zero. Set $$ \Delta = 4ab - t^2 $$ and $$ \delta = 4 a c - s^2. $$ Then $$ 4 a \Delta g = \Delta \left(2ax + ty + sz ...


0

What if you would take a point $x\in B$. Then consider a line perpendicular to your ellipse $B$, through $x$. Now if this line intersects $A$ in a point $y_1$ and possibly $y_2$, where we suppose $y_1$ and $y_2$ to be outside $B$. Take then $d_x = max(d(x,y_1),d(x,y_2))$, where $d(x,y)$ is the distance between $x$ and $y$. Then we do this for all $x\in B$ ...


0

If you have a conic section in the form $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ then the determinants $$\delta= \begin{vmatrix} a & h \\ h & b \end{vmatrix}=ab-h^2$$ and $$\Delta= \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}= abc+2fgh-af^2-bg^2-ch^2$$ are invariants - which mean that they do not change ...


2

This solution is closely related to the Lorentz transformation solution I posted earlier, but is perhaps closer to the heart of the matter. Apologies for posting multiple solutions, but I hope each adds something interestingly new. Assume without loss of generality that the side-length of the glass is $1$ and the apex a right angle, with $x$ and $y$ axes ...


2

WARNING: this answer uses the new parameterization of points introduced by the OP: \begin{eqnarray} A & = & \left( r\cos { \alpha } ,t\sin { \alpha } \right) \\ { A }' & = & \left( r\cos { \alpha } +\sqrt { 2 } \cos { \left( \frac { \pi }{ 4 } +\frac { \alpha }{ 2 } \right) } ,t\sin { \alpha } +\sqrt { 2 } \sin { \left( \frac { ...


0

This is not an answer but an illustration of the correct ellipse based on N74's answer. It seems the OP's ellipse was in fact right on target. Below is the ellipse I get when the double of the angle given by N74 is used. The area $A = 11525.27$.


0

Here, I just cover the key part: Expressing $$a+b+c+d=R$$ as a polynomial equation in $R$ and $(A,B,C,D)=(a^2,b^2,c^2,d^2)$. Then you can plug in the distance squares $A=x^2+y^2$ etc and do all the messy expansions. Let $S_k = a^k+b^k+c^k+d^k$. We note $$\begin{align} S_1 &= R & S_{2k} &= A^k+B^k+C^k+D^k \end{align}$$ And for every ...


1

The first conclusion you have to make is that the problem is symmetric on both the $x$ and the $y$ axis, so the equation of the ellipse we want to find is of the form $$ {x^2 \over a^2}+{y^2 \over b^2}=1.$$ To find a solution to this problem we can, then, force the ellipse to be tangent to one of the circles, and the ellipse will be tangent to all the ...


1

This is inspired by remarks several people made in thinking about the martini puzzle: the glass is a light cone, and tipping is a Lorentz transformation! I want to show that at least the latter is literally true. A Lorentz transformation maps a coordinate system into another, based on one parameter $v$, normally thought as the relative velocity of two ...


2

As a start, let us first derive the formula when the unit square is centered at origin. Let $a = x^2+y^2+\frac12$, the distance to vertex $v$ is: $$\begin{cases} r_{++} &= \sqrt{a - x - y}, & v = (+\frac12,+\frac12)\\ r_{+-} &= \sqrt{a - x + y}, & v = (+\frac12,-\frac12)\\ r_{-+} &= \sqrt{a + x - y}, & v = (-\frac12,+\frac12)\\ ...


0

If by "intersecting at point" you mean external tangency then the answer to your question is "No". You can refer to the first diagram on this link below See a diagrammatic example


0

It's not clear exactly what you did. If you were doing something like Y = (-0.5(X+1)² + 9)(2.7>X)(-6.9<X) then this will cause the value of Y to be zero when X is outside the selected range (because one of the two operators > or < will produce the value $0$), but it will not make the value of Y be undefined, which is what you want. You might ...


1

We assume the glass cone has height and radius $1$ (hence side length $\sqrt{2}$) without loss of generality, because any other form factor can be obtained by uniform scaling in the direction of the glass's axis, which preserves ratios of volumes (liquid to glass in both upright and tipped configurations), and of lengths of parallel line segments ($p$ and ...


1

Part (iii): As pointed out by @mathlove, $$p=-1/t^3$$. Part (iv): Coordinates of $M$ are given by $$x=\frac 12 \left(ct-\frac c{t^3}\right)\qquad\Rightarrow \frac {2x}c=t-\frac 1{t^3}=X\\ y=\frac 12 \left(\frac ct-ct^3\right)\qquad\Rightarrow \frac {2y}c=\frac 1t-t^3=Y\\\\ \frac XY=-\frac 1{t^2}\\ XY=2-t^4-\frac 1{t^4}=-\left(t^2-\frac ...


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EDIT: This picture illustrated the basic idea of the proof below. (Although it is written using the parametrization of ellipse, it roughly corresponds to the picture.) I would say that the idea behind almagest's answer is similar. The basic idea can be explained like this: Relation between ellipse and circle is that we scaled the coordinate along one of ...


2

Without loss of generality $a>b$. Take an inscribed rectangle with two sides gradient $k\ne0$ and two sides gradient $-\frac{1}{k}\ne0$. Shrink along the $x$-axis by a factor $\frac{b}{a}$. The two sides gradient $k$ now have gradient $\frac{ka}{b}$ and the two sides gradient $-\frac{1}{k}$ now have gradient $-\frac{a}{kb}$. So the rectangle is now a ...


0

Rotate the ellipse by $ \tan^{-1}\frac{ B} { A-C } $ to bring them to form where axes are parallel to axes. To get a feel, plotting them parametrically will help.The ellipse plots with $$ x = H + A \cos t ,\,y= K + B \sin t ;$$ Next plot circles by varying $ h,k,a$ $$ x = h + a \cos u, \, y=k + a \sin u $$ along with ellipse to see if and where they ...


0

Suppose we have a rectangle which has all four vertices on the ellipse. If $C$ is the center of the rectangle, then all vertices of the rectangle are in the same distance from $C$. If we choose coordinate system with origin in the point $C$, we get the equation $$x^2+y^2=r^2.$$ We also want to express the condition that the vertices belong to the ellipse. So ...


1

(iii) You have an error here. That the normal to H at the point $T$ meets $H$ again at $P(cp , \frac{c}{p})$ means that the normal to $H$ at the point $T$ is the line $TP$. The gradient of the line $TP$ is given by $$\frac{c/p-c/t}{cp-ct}=\frac{-(p-t)}{(p-t)pt}=-\frac{1}{pt}$$ Since this is equal to $t^2$, we have ...


1

1) Consider the point as a vector $$ p=\begin{bmatrix} x \\ y \\ \end{bmatrix} $$ 2) Consider the center of the ellipse as $$ c=\begin{bmatrix} h \\ k \\ \end{bmatrix} $$ 3) Subtract the center of the ellipse $$ p_{centered}= p - c $$ 4) Create a whitening matrix $$ W = ...


5

Diagram below shows side and aerial views of upright and tilted martini glass. For the upright glass, the shape of the martini liquid is an inverted right circular cone, with height $p\sin\theta$ and circular base of area $A_1$. Volume of martini is given by: $$\begin{align} V&=\frac 13\cdot A_1 \cdot p\cos\theta\\ &=\frac 13\cdot \pi ...


3

This is not an answer to the stated question, just an outline and an example of how to compute the envelope (and thus area) numerically, pretty efficiently. The code seems to work, but it is not tested, or nowhere near optimal in the algorithmic level; it is just a rough initial sketch to explore the problem at hand. The sofa is symmetric with respect to ...


6

Consider this diagram, representing a side view of the martini glass with "tipped" so that the beverage reaches a maximum distance $x := |\overline{OX}|$ and minimum distance $y := |\overline{OY}|$ from cone vertex $O$: Let $O^{\prime\prime}$ be the midpoint of $\overline{XY}$, so that it is the center of the represented ellipse; thus, ...


3

On the surface of the cone, $\frac rz=\tan\theta$. On the planar surface of the fluid, using the point-slope formula for a line $$\frac{z-h}{r\cos\psi-h\tan\theta}=\tan\phi$$ Where $\phi$ is the angle of tilt of the glass and $\psi$ will be the azimuthal angle. $\psi=0$ at point $E$ in the diagram. On the meniscus, $$z=\frac ...


0

At the risk of sounding grumpy, I don't unless I have to. I try to write my conic sections in the closest form to a description of its properties I can get away with. I don't try to understand conic properties as a result of cartesian coordinates, because that's backward. I do keep track of what methods are necessary to distinguish and retrieve ...


1

My experience is so far this: it is difficult to read, and worth every effort. Advice: Do not delay. Read it. It is rough going at first, so make a plan. I recommend forcing yourself through the first 16 propositions, building the problems in a 3D-capable Geometry program. This gets us from a point and a line to the identification of the sections, each ...


1

I have a complete solution which works fine for me but should undergo a bit more testing. I made it because Geogebra is terrible at finding roots. My approach will benefit only modestly from parallelization. I am not convinced matrices are an improvement, but for a graphics library it is surely worth checking. My experience is this: symbolically, they ...


0

No there isn't because the sphere is optimal in relation to the ratio of circumference to area. You can check this simply by writing the equation of the circumference divided by the area and differentiating with respect to the width to see that the minimum occurs when the width and length are equal. Although technically a circle is an ellipse so the actual ...


-1

I cannot give you a deep rigourous argument, but surface tension is the reason a soap-bell is spherical so... A circle has a specific eccentricity, and you choose another eccentricity while the area is fixed, there are no more parameters left to adjust for the perimeter.


1

No, because the isoperimetric quotient $A/P^2$ is smaller for a non-circular ellipse than for any circle.


0

Let the drawing plane pass through the vertex of the double cone. And let us agree to call such a section "degenerate". I say, that the resulting section is a Pair of Intersecting Lines, if any part of the plane passes inside the cone. a Line, if the plane touches (is tangent to) the conic surface. a Point, if the only point of contact is the vertex. ...


1

There is some information on a more complete classification of conic sections on this page. With these notations, your equation: $$(E) \quad x^2+xy+ky^2+6x+10=0$$ has the associated matrix $A_E$ given by: $$A_E = \begin{pmatrix}1 &0.5&3\\0.5&k&0\\3&0&10\end{pmatrix}$$ Following the classification rules described in the article, you ...


0

Three points determine a parabola with axes parallel to either x- or y-. One more point or condition is lacking here for full a solution. But upto a constant, $ y = a \cdot x \cdot (5-x) $ are all parabolas with variable $a$ passing through the given points.


0

You do not have enough information to get an explicit value for a. As you said, your function will have the form: $$y=ax^2+bx+c$$ Plugging in $y=0$ and $x=0$ tells us that $0=c$. So now we have the new equation: $$y=ax^2+bx$$ Plugging in the second point we see that $0=25a+5b\Rightarrow 5a=-b$ yielding now $$y=ax^2-5ax$$ And now we need to use your last ...


1

If you were clever you would say, I know 2 roots, therefore: $y = a(x-0)(x-5)$ now, what values can $a$ take on? If you were more brute force about it. $f(x) = ax^2 + bx + c\\ f(0) = c = 0\\ f(5) = 25a + 5b = 0\\ b = -5a\\ f(x) = ax^2 -5a x\\ $ And you still need to make sure that a is in the right set of values. but you have chosen $y = a (x-h)^2 + v$ ...


2

If we set $P_i=(x_i,y_i)=\left(3\cos\theta_i,2\sin\theta_i\right)$ we have $$\vartheta_0+\frac{\pi(i-1)}{n}=\arctan\frac{y_1}{x_i}=\arctan\left(\frac{2}{3}\tan\theta_i\right)$$ from which: $$ \tan\theta_i = \frac{3}{2}\tan\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) $$ and $$ OP_i^2 = 9\cos^2\theta_i+4\sin^2\theta_i = 4+5\cos^2\theta_i=\frac{9+4 ...


2

Better to set up a system of two equations: $$ 3\cos\theta_2=r_2\cos(\pi/n) \quad\hbox{and}\quad 2\sin\theta_2=r_2\sin(\pi/n), $$ where $r_2=OP_2$. From that you get, after eliminating $\theta_2$: $$ {1\over r_2^2}={\cos^2(\pi/n)\over9}+{\sin^2(\pi/n)\over4}, $$ and in general: $$ {1\over r_{k+1}^2}={\cos^2(k\pi/n)\over9}+{\sin^2(k\pi/n)\over4}= ...


1

I use some of the same properties, particularly the "optical" property, as the other responders here, but in a somewhat different way, so some of the same calculations will appear, but in a different guise. (This seems closest to hypergeometric's approach.) There is a "similarity" property we can apply. For the "upward-opening" parabola with vertex at the ...


3

The equation is obviously $$ \sqrt{x^2+y^2}+\sqrt{x^2+(y-1)^2}+\sqrt{(x-1)^2+y^2}+\sqrt{(x-1)^2+(y-1)^2} = R $$ (for $R\geq 2\sqrt{2}$) that is the equation of an algebraic curve of degree $10$. For large values of $R$, such curve is closer and closer to the circle centered at $\left(\frac{1}{2},\frac{1}{2}\right)$ with radius $\frac{R}{4}$.


1

Eliminating $C_i$ gives $$x^3=y^2\qquad\blacksquare$$ There are two lines because the equation is equivalent to $$x^3=(\pm y)^2$$ i.e. for each value of $x$ there are two possible values ($\pm$) of $y$.


0

An ellipse centred at $(4,3)$ with major and minor axes parallel to the $x-$ and $y-$ axes respectively has the following equation: $$\frac {(x-4)^2}{a^2}+\frac{(y-3)^2}{b^2}=1\qquad (a>b>0)\qquad\cdots(1)$$ Applying the rotation matrix for $\theta=-\pi/4$ and simplifying: $$\frac{(x+y-7)^2}{2a^2}+\frac{(x-y-1)^2}{2b^2}=1\qquad\cdots(2)$$ ...


1

if you shift the origin to eliminate the linear terms the equation becomes $$ rMr^T = C $$ where $r$ is the 2-vector $(x,y)$ and $M$ is the matrix $$M = \begin{pmatrix} a & h \\ h & b \end{pmatrix} $$ and $C$ is a constant. If the axes are rotated through an angle $\theta$ the matrix is diagonalized when the off-diagonal terms in $$ ...


1

If we rotate the axes through $\theta$ to new axes $x',y'$ we have $x=x'\cos\theta-y'\sin\theta,y=x'\sin\theta+y'\cos\theta$. Hence the coefficient of $x'y'$ becomes $-2a\cos\theta\sin\theta$ (from the $ax^2$ term) plus $2b\cos\theta\sin\theta$ (from the $by^2$ term) plus $2h(\cos^2\theta-\sin^2\theta)$ (from the $2hxy$ term). So the $x'y'$ term has zero ...


0

Note that for clockwise rotation by $45^{\circ}$, \begin{align*} \begin{pmatrix} x' \\ y' \end{pmatrix} &= \begin{pmatrix} \cos (-45^{\circ}) & -\sin (-45^{\circ}) \\ \sin (-45^{\circ}) & \cos (-45^{\circ}) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ &= \begin{pmatrix} \cos 45^{\circ} & \sin 45^{\circ} \\ ...


6

The diagram shows a point $P$ between the arms of a hyperbola. It also shows how the rays emanating from $P$ fall into four regions (denoted "North", "South", "East", "West") bounded by $\overrightarrow{PA}$, $\overrightarrow{PB}$, $\overrightarrow{PC}$, $\overrightarrow{PD}$: Rays in the North and South regions never hit the hyperbola at all. Rays in the ...


1

(see figure) @wojowu @Blue @Shreyash Chaudhari Here is a mathematical explanation, using projective geometry. Let us consider the case of an ellipse : from a given point outside an ellipse, you can draw two tangents (only). If you have done projective geometry, you know that one can transform any conic section into any other one, in particular, one can ...


0

Thank you. I found a much easier way to get the properties of the ellipse using the Rytz's construction method.


1

Your solution does not work. First, $m$ is always different from $0$ because $H(\mathbb Q)$ contains the cyclic subgroup $\{(1,0),(-1,0)\}$. Second, what does it mean that "$\mathbb Q(\sqrt{A})^*$ has no linearly independent elements"? Because actually $\mathbb Q(\sqrt{A})^*$ contains a lot of linearly independent elements! To prove the claim, one can ...


0

In the interest of avoiding confusion, I’m going to use $(l,t)$ instead of $(x,y)$ for the upper-left corner of the bounding box. The question you reference gives you the equation of an ellipse centered on the origin. If its center is elsewhere, but it’s still aligned with the coordinate axes, then getting its equation is a simple matter of translating the ...


2

You might know that the equation of the common chord of two circle whose standard equations are $S_1=0$ and $S_2=0$ , is $S_1=S_2$ . The required circle bisects the circumference of all the other 3 circles, so its common chord with these circles should pass through their respective centres. So, we have: $$c=-5\tag{i}$$ $$8g+6f+c=-32-18+10\tag{ii}$$ ...



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